
fe^ T h356 

Book 



l~l^ 



Gopjiigiit]^". 



COPYRIGHT DEPOSIT. 



The D. Van Nostrand Company 

intend this book to be sold to the 
Public at the advertised price, and 
supply it to the Trade on terms which 
will not allow of reduction. 



THE ELEMENTS OF JilECHANICS 
OF xMATERIALS 

A TEXT FOR 
STUDENTS IN ENGINEERING COURSES 



BY 



C. E. HOUGHTON, A.B., M.M.E. 

ASSOCIATE PROFESSOR OF MECHANICAL ENGINEERING 
NEW YORK UNIVERSITY 




NEW YORK 
D. VAN NOSTRAND COMPANY 

1909 



^^s* 



COPYRIGHT, 1909, 
BY D. VAN NOSTRAND COMPANY. 



Norbjootj iSrJss : 

Set up and electrotypeil by J. S. Gushing Company, 

Norwood, Mass., U.S.A. 




<£) 



LIBRARY of CONGRESS 
Two CoDies Weceived 

JUL / l^^^ 

opyriiint Entry ^ 

^ 4 "^ ; ?. 4 




PREFACE 

This is not a treatise on the Mechanics of Materials. 
The efforts of Merriman, Burr, Lanza, and others cover 
the field so thoroughly that there is no present need of 
such a work. 

It is designed to be an elementary text-book for students 
in the engineering courses in colleges and universities, 
where the time allotted to the subject does not exceed 
three or four recitations per week, for one half year, and 
where the course is preceded by college courses in mathe- 
matics, through integral calculus, mechanics, and physics. 

The extreme mathematical treatment of the subject has 
been avoided, but where the use of higher mathematics 
leads to clearness they have been freely used. 

As it is intended as a text-book, the general cases are 
discussed fully, leaving the student to derive the formulas 
for special cases as part of the regular problem work. 

At the end of each chapter there are review questions 
covering the more important parts of the subjects dis- 
cussed and problems illustrating the same. The solution 
of one problem of each type has been given to show the 
application of the general formulas. 

The appendix contains tables giving the values of the 
engineering constants of materials and the formulas com- 
monly used in design, in addition to the tables usually 
found in books of this character. 

The notation has been made uniform Avith that of Mer- 
riman's works, so that his more complete treatise on the 
subject may be conveniently used as a reference book. 

New York, January, 1909. 

iii 



TABLE OF CONTENTS 



CHAPTER I 



stresses 



Applied Mechanics 

ARTICLE 

1. Forces in structures 

2. Axial forces .... 

3. A bar 

4. Internal forces 

5. Tensile or compressive stresses 

6. Unit stress .... 

7. Maximum tensile or compressive 

8. Shearing stresses . 

9. External and internal forces . 

10. Deformation of elastic bodies 

11. Unit deformations . 

12. Modulus or coefficient of elasticity 

13. The elastic limit 

14. Ultimate strength . 

15. Resilience .... 

16. Ductility 

17. Elastic resilience 

18. Use of formulas 

19. Constants of materials . 

20. Units 

21. Working stresses ; factors of safety 

22. Accuracy of calculations 
Examination questions . 
Problems .... 



1 

3 

3 

3 

4 

4 

5 

5 

6 

7 

7 

8 

8 

9 

10 

11 

12 

12 

13 

14 

14 

15 

16 

18 



CHAPTER II 

Applications 

23. Bars of uniform strength 

24. Thin pipes, cylinders, and spheres 

V 



22 

25 



VI 



TABLE OF CONTEXTS 



•Jo. 
26. 
27. 
28. 
29. 
30. 
31. 
32. 
33. 



Thick pipes 

Riveted joints 

Tension in plates . 

Shear on rivets 

Conijjression on rivets or plates 

General case of a riveted joint 

Kinds of riveted joints . 

Efficiency of a riveted joint . 

Stresses due to change of temperature 

Problems 



PAGE 

27 
30 
30 
32 
33 
33 
35 
37 
39 
40 



34. 
35. 
36. 
37. 
38. 
39. 
40. 
41. 
42. 
43. 
44. 
4.5. 
46. 
47. 
48. 
49. 
50. 
51. 



CHAPTER III 

Beams 

Kinds of beams 

Reactions at the supports 

Uniform and concentrated loads 

Vertical shear 

Bending moment . 

Resisting shear 

Resisting moment . 

Use of formula 

Shear and moment diagrams . 

Shear diagrams 

^Moment diagrams . 

The relation between the vertical shear and bending moment 

Relative strengths of simple and cantilever beams 

Overhanging beams 



Beams of uniform streng 
Moving loads . 
Use of formula 
Examination questions 
Problems 



th 



46 
47 
47 
49 
50 
51 
52 
55 
58 
59 
60 
63 
64 
65 
66 
67 
69 
70 



CHAPTER lY 
Torsion 



52. Derivation of formula 

53. Modulus of section 

54. Square sections 



83 

86 
86 



TABLE OF CONTEXTS 



Yll 



ARTICLE 

55. Illustrations .... 

56. Twist of shafts 

57. Relative strengths and stiffness 

58. Horse power of shafts . 

59. Shaft couplings 

60. Modulus of rupture for torsion 

61. Helical springs 
Examination questions . 
Problems .... 



PAGE 

87 
88 
89 
89 
90 
92 
93 
94 
95 



CHAPTER V 
The Elastic Curve 



62. Definition .... 

63. The equation of the elastic curve 

64. Deflection of beams 

65. Fixed or restrained beams 

66. Beams fixed at both ends 

67. Continuous beams . 
Examination questions . 
Problems 



99 
99 
102 
105 
107 
109 
lU 
116 



CHAPTER VI 

Long Columns 

68. Stresses in long columns 

69. Euler's formula for long columns . 

70. Columns with round or pin ends . 

71. Columns with square, flat, or fixed ends 

72. Columns with round and square ends . 

73. Rankine's formula for long columns 

74. Applications 

Examination questions . . . . 
Problems 



123 
124 
125 
126 
128 
131 
135 
137 
139 



CHAPTER VII 
Combined Stresses 

75. Stresses due to force ..... 

76. Tension or compression combined with bending 



143 
143 



Vlll 



TABLE OF COXTENTS 



ARTICLE 












PAOE 


77. Roof rafters 147 


78. Eccentric axial loads 












. 149 


79. Shear and axial stress 












. 149 


80. Maximum stresses . 












. 150 


81. Horizontal shear in beams 












. 152 


82. jNIaximum stresses in beams 












. 1.56 


Examination questions . 












« 158 


Problems 












. 160 



CHAPTER Vin 



Compound Bars and Beams 

83. Definition 

84. Compound columns, alteinate layers 

85. Compound columns, longitudinal layers 

86. Compound beams ..... 

87. Reenforced concrete beams 

88. Straight line formula for reenforced concrete 
Examination questions .... 
Problems 



Tables, Explanation of . 

Table 1. Notation .... 

Table 2. Fundamental formulas . 

Table 3. Derived formulas . 

Table 4. Properties of beams 

Table 5. Average constants of materials 

Table 6. Properties of sections . 



beams 



163 
163 
163 
165 
166 
166 
173 
174 

176 
176 
178 
179 
182 
183 
184 



MECHANICS OF MATERIALS 



CHAPTER I 
APPLIED MECHANICS 

Article 1. Forces in Structures. 

One of the problems that confronts the engineer called 
upon to design any machine or structure is to so propor- 
tion the various parts that they will resist the forces that 
act on them. 

To do this, he must apply the laws of mechanics to the 
forces to be resisted, and study the action of the materials 
under the same forces. 

This application of mechanics may be termed Applied 
Mechanics or the Mechanics of Materials. 

If we consider any structure or any member of the 
structure to be at rest, according to the laws of mechanics 
the forces that act on the structure or member must be in 
equilibrium. The various parts of a machine often have 
relative motion, but by introducing a force equal and op- 
posite to the force which produces that motion, the forces 
that act on the member of a machine may also be treated 
as a system of forces in equilibrium. 

In the more extended treatment of this subject the 
forces are taken as acting in different planes. The simpler 
theory that treats the forces as coplanar is the one that 
will be used here. 

1 



2 MECHANICS OF MATERIALS 

Since each member must be designed separately, if the 
forces that act on any member are determined, we will find 
that the forces may be resolved into : 

Forces tending to lengthen the member, 
Forces tending to shorten the member, 
Forces tending to bend the member. 
Forces tending to shear the member. 

Imagine any body, acted on by a system of forces in 
equilibrium, denoting the sum of the components of the 
forces parallel to some line as the X forces, and those per- 
pendicular to the same line as the 1^ forces; the sum of the 
X foi'ces is zero and the sum of the Y forces is zero. If 
we take as the line of reference the axis of the body pass- 
ing through the center of gravity of the body, the result- 
ant of the X forces will be a couple, unless the lines of 
action of the forces of the couple coincide with the axis. 

Each force of this couple may be replaced by a single 
force of equal magnitude, acting in the line of the axis and 
a couple whose moment is the moment of the force about a 
point in the axis. The X forces acting in the line of the 
axis will tend to either lengthen or shorten the member 
in the line of the axis, and the couple as well as the Y 
forces will tend to bend the member. 

If the member be cut by a plane perpendicular to the 
axis, the Y forces on either side of the section will be 
opposite in sign, and in general tend to slide one part of 
the member, relative to the other, along the plane of the 
section. Any of the resultants may be zero, and in that 
case there would be no tendency to deformation in that 
line. 

The force acting on any member is always transmitted 
by a surface of finite area, but by considering that each 



APPLIED MECHANICS 



elementary area of the surfaces in contact transmits the 
same amount of pressure, we may use the resultant of 
these elementary pressures passing through the center of 
gravity of the areas in contact, as the force applied. 

Art. 2. Axial Forces. 

When the lines of action of the acting forces lie in 
the axis of the body, passing through the centers of 
gravity of all sections perpendicular to the axis, the forces 

are called Axial 
Forces, and their 
effect is to 
either lengthen 
or shorten the 
member. 



V 



V 



\ 



a 



\^ 




Fig. 4. 



a 

Art. 3. A Bar. 

The member 
may have any 
shape whatever, 
JC the simplest be- 
ing a prismatic 
or cylindrical 
form, where any section perpendicular 
to the axis has the same shape and area. 
This form will be called a Ba7\ 

Art. 4. Internal Forces or Stresses. 

Let Fig. 4 represent a bar under the 
action of the axial forces P and P. 
Suppose the bar to be cut by any plane 
through the axis into the segments a 
and 5, and consider the 



segment 



a. 



4 MECHANICS OF MATERIALS 

This segment is acted on by the external force P, and as a 
part of the whole bar it is in equilibrium ; hence there must 
be forces acting in the plane section X-X^ whose resultant 
acts in the same line, and is equal and opposite to the force 
P. As the same is true of the segment 6, the internal 
forces, called stresses, acting between a and 5, hold the 
segments a and h in equilibrium against the external forces. 
Therefore, in any plane section of the bar there exists 
a pair of equal and opposite forces or stresses, each of 
which are induced by and resist the external forces. In 
general, the stresses may be resolved into components 
parallel and perpendicular to the plane section. The 
components parallel to the plane of the section prevent 
the sliding of the segments along the plane, and are 
termed Shearing Sti'esses^ while those perpendicular to the 
plane are called either Tensile or Compressive Stresses^ 
depending on whether they tend to extend or compress 
the particles on which they act. 

Art. 5. Tensile or Compressive Stresses. 

When the external forces are axial, and the section per- 
pendicular to the axis, the stresses can have no component 
parallel to the plane of the section; hence axial forces can 
produce only tensile or compressive stresses in planes per- 
pendicular to the axis. The plane dividing the bar into 
the segments a and h was any plane ; hence the reasoning 
holds true for all such planes, and there are only tensile or 
compressive stresses equal to the external force P, in all 
sections perpendicular to the axis. 

Art. 6. Unit Stress. 

Since the force P is the resultant of all the equal unit 
pressures on the areas in contact, it is reasonable to assume 



APPLIED MECHANICS 5 

that the stresses on each unit of area of the plane section 
are also equal, and if iS equals the sum of the stresses 
acting on each unit of area of the plane section, and A is 
the area of that section, then 

P = AS. («) 

S is termed the unit stress, and is the resisting force per 
square unit of area ; hence S must be expressed in the 
same units as -P and A. 

Art. 7. Maximum Tensile or Compressive Stresses. 

When the cutting plane is not perpendicular to the 
axis, the resultant stress may be resolved into components 
parallel and perpendicular to the plane of the section, 
those perpendicular being either tensile or compressive 
stresses, while those parallel are shearing stresses. 

As neither component can equal the resultant, it is evi- 
dent that the maximum tensile or compressive stresses 
will be found in a section perpendicular to the axis. In 
such a section there are no shearing stresses, and when 
the bar has a uniform section area A, the formula (a) will 
determine the maximum tensile or compressive unit stress 
induced by the axial force P. If the areas of all sections 
j)erpendicular to the axis are not equal, the greatest unit 
stress will be found where the section area is the least, 
and the value of A to be used in formula («) is the area 
of the least section. 

Art. 8. Shearing Stresses. 

When the external forces act in adjacent parallel lines, 
since the stresses can have no component perpendicular to 
the line of action of the forces, the stress in a section 
parallel to the line of action of the forces must be a 



MECHANICS OF MATERIALS 



Shearing Stress^ as the external forces tend to slide the two 
sections of the bar along the plane of the section. (Fig. 2. ) 
Assuminor that the stress is uniformly distributed over 
the section, and that Sg is the unit stress in shear, then 
formula (a) P = AS^, where A is the area of the section 




Fig. 8. 

parallel to the line of the forces and P the forces pro- 
ducing the shear, will always give the relation between 
the external forces and the maximum unit shearing stress 
in the section. 

Art. 9. External and Internal Forces. 

The external forces on any member of a machine or 
structure are the weights or loads that member has to 
support and the pressure it receives from the adjacent 
members, while the internal forces are those that transmit 
the external force from element to element through the 
member. These latter forces are stresses, and the inter- 
nal force per unit of area is the Unit Stress, and will be 
designated by the letter S. with a subscript t, c, or s, as 
the stress is tension, compression, or shear. 

The formula P = AS is a general one and applies to all 
cases where the stress is uniformly distributed over the 
area of any plane section A. and aS', the kind of unit 
stress. 



APPLIED MECHANICS 7 

Art. 10. Deformation of Elastic Bodies. 

In the study of mechanics, the forces were assumed to 
act on rigid bodies; tliat is, on bodies whose shape was not 
altered by the application of the force. As there are no 
rigid bodies in nature, every force applied produces some 
deformation or change of shape. This fact, however, does 
not prevent the application of the laws of mechanics to 
elastic bodies under the action of force after the deforma- 
tion has taken place, since equilibrium must exist at that 
time. 

Art. 11. Unit Deformations. 

Consider a bar, I units in length, and A square units in 
section, under the action of an axial force P. From equa- 

P 

tion (a), aS' = —-. aS'Is constant, since P and A are constant 
A 

for all sections perpendicular to the axis, or each square 
unit of every section is acted on by a force S. 

Suppose the bar to be divided into bars, each one unit 
in section and I units long, then each of these bars is acted 
on by a force S. The change in the length Z, that may 
take place under the force aS', being e, since all particles 
are under the same force, the change in the length must 
be equal to the length I, multiplied by the unit load S^ 
and some number which depends on the nature of the 

material. Calling this number— ,. the value of e must be 

E 

e ='—. This may be written - = '—. Letting - = e equal 
the change in the length of a bar one unit in length, or the 

unit deformation, it follows that e = — or 

E 

jji _S^_ unit stress ^tn 

6 unit deformation 



8 MFXHANICS OF MATERIALS 

Art. 12. Modulus or Coefficient of Elasticity. 

If the value of S does not exceed a critical value which 
varies for different materials, ^ will be constant for all 
values of S, and this constant is called the Modulus or Co- 
efficient of Elasticity. This constant is the value of the 
ratio of the unit stress to the unit deformation, as may be 
seen by the inspection of formula (6). 

PI P 

Equation (5) may be written E = — -, since S = — and 

€ = -, and as A and I are constants for any bar under axial 
forces, E will be constant when e varies directly with P. 

Art. 13. Elastic Limit. 

If a bar length I is subjected to a small axial force P, it 
is observed that the length has changed a certain small 
amount. If P^ is twice P, experiment shows that the 
change in length is twice that due to P, and if P„ is n 
times P, the change in the length is n times that due to 
P, provided that P„ does not exceed a certain limiting 
value which varies for each material and bar. That is, 
within that limit the change in length of any bar is pro- 
portional to the external force applied. If P„ is the 
limiting value for a given bar, then the corresponding 
value of S as derived from formula (a) must be the 
limiting value of the unit stress. This value of aS', being 
a unit stress, is independent of the dimensions of the 
bar and depends only on the material of the bar, is 
called the Elastic Limit of the material or the limit of 
elasticity. 

The elastic limit of a material may be defined as the 
unit stress for which the deformations cease to be propor- 
tional to the applied force. 



APPLIED MECHANICS 9 

To determine this value of S for any material, axial loads 
are applied to a bar of the material, the loads being applied 
in small equal increments, and the change in length due to 
each increment of load measured. For any total load less 
than the load producing a stress equal to the elastic limit, 
the last increment of load should produce the same change 
in length as any previous increment. Therefore, when an 
increase in the change in length for any equal increment 
of load is noted, the elastic limit has been passed. It will 
be noticed that the exact value of elastic limit depends on 
the accuracy with which the loads and especially the de- 
formations are measured. 

It has also been observed that if the stress in any bar is 
less than the elastic limit, the bar will return to its origi- 
nal length when the load is removed, and if the stress is 
slightly above the elastic limit, there will be some perma- 
nent change in the length of the bar. The unit stress at 
which this yielding takes place is called the Yield Point 
or the Commercial Elastic Limit. The latter term comes 
from the commercial practice of determining the elastic 
limit by the drop of the beam in the testing machine. 

While the yield point or commercial elastic limit 
is from 3 to 5 % higher than the true elastic limit, the ease 
with which the latter value may be determined and the 
fact that the allowable value of S for any engineering 
structure rarely exceeds one half of the elastic limit, com- 
bine to make it the one in general use. 

Art. 14. Ultimate Strength. 

After the elastic limit is reached, the change in length 
increases more and more rapidly as the loads are increased, 
and finally a load is reached that causes the bar to rupture. 



10 



MECHANICS OF MATERIALS 



If P is the load that causes rupture and A is the original 
area of the bar, then the value of aS' obtained from formula 
(a) is called the Ultimate Strength of the material. 

Art. 15. Resilience. 

If a bar of wrought iron, whose section area is A^ and 
the length I as measured between two punch marks on 
the bar, is placed in a testing machine, the tensile loads 
and their corresponding extensions being measured and 
plotted to scale on section paper, the result would be a. 
diagram similar to that in Fig. 15. 









h 


•^ 

^ 

^ 






^^ ' ^ 


§ 
•^ 




^^-"^^ 




^ 




/l 




to 

CO 
CO 


xl 


1 




1 


l\ 











y J Change in length, or unif-cjeformation. 

Fig. 15. 



e 



Since P = ^aS', the ordinates representing loads may,, 
by a change of scale, represent unit stresses, and the ab- 
scissa representing total elongations may also represent 
unit elongations. Such a diagram is called a Stress-strain 
diagram. The use of the word "strain" gives it the 
meaning of "unit deformation." Authorities do not agree 
on the use of the word, some giving it the meaning as 
above, while others use it to mean load. On account of 



APPLIED MECHANICS 11 

this ambiguity, the term "unit deformation" will be used 
in its place. The unit stress for any load is obtained by 
dividing the load by the original area, and the unit elon- 
gations by dividing the elongation for any load by the 
original length I. 

The point a on the curve is the elastic limit, and b 
is the ultimate strength of the material. After a load 
corresponding to h is reached, the bar begins to reduce at 
some point very rapidly, finally breaking at a load less 
than the load at h. This load is called the Breaking Load^ 
and has but little significance in engineering work. 

Since one coordinate represents force, and the other 
space, the area o a h c e^ when measured in the proper 
units, is the work done in breaking a bar of unit volume. 
If we define the Ultimate Resilienee as the work done in 
breaking a bar of unit volume, the area represents that 
quantity. 

Art. 16. Ductility. 

As the whole curve is rarely ever determined, the Duc- 
tility^ a term that is defined by its method of calculation 
and is proportional to the ultimate resilience, is generally 
used in its place. 

The ductility of any material is calculated as follows: 
After the bar has been broken by a tensile load, the pieces 
are removed from the testing machine and the broken ends 
placed together. The distance between the original punch 

marks, being ?, has now increased tol-\-p', then ^ is called 

the Ductility^ and is usually stated as a percentage of the 
original length. 



12 MECHANICS OF MATERIALS 

Art. 17. Elastic Resilience. 

The area oaf is easily calculated, as it is the area of 
a triangle, and when measured in work units is called the 
Elastic Resilience. This work is evidently 

the unit stress at 6g x the unit deformation at a 

2 

Calling this value ^, it is the elastic resilience of the 

material, or the work done in raising the unit stress in a 

bar of unit volume from to the elastic limit. 

The same reasoning is true for any stress less than the 

elastic limit, giving a method of calculating the work 

necessary to produce any given stress less than the elastic 

limit in a bar of unit volume. If >S' is any stress less than 

the elastic limit, and e the unit deformation at that stress, 

then A: = J iS'e is the work done per unit of volume, and 

the work done on any bar whose volume is V is V 

times that quantity. Taking the work done on any 

P e 

bar as ^ Se x V] substituting for S its value -— , for e, -^ 

and for FJ Al^ the expression for the work done on any 

1 Pe 

bar reduces to IC=- — — which is in terms of the total 

2 Al 

load and deformation, and is true when P is less than the 
load, causing a stress within the elastic limit. 

Art. 18. Use of Formulas. 

Whenever any stress can be assumed to be uniformly 
distributed over the area of any section of a bar, formula 
(a), P = AS^ gives the relation between the load and the 
stresses on that area, and any two of these quantities 
involved being given, the other may be found. 

The relation between the load and the deformation for 



APPLIED MECHANICS 13 

S PI . . 

such cases is given by U = — = —— , which is true when 

e Ae 

S = — is less than the elastic limit, and is applicable to 

A. 
all problems involving the deformations of a bar under 

axial loads. 

Also k = i Se and K= - — - can be used under the 
^ 2 Al 

same conditions when the data for the problem include a 
consideration of the work done in deforming a bar. When 
the load is axial and the stress may be taken as uniformly 
distributed over the area of any section, these three equa- 
tions furnish the means for the design and investigations 
of the strength of all members of a structure or machine. 

Writing (a) S = ^, (5) U=-, and ^= J Se, it will be 
A e " 

noticed that the equations are simply the algebraic ex- 
pressions of the definitions of Unit Stress, Modulus of 
Elasticity, and the Unit Resilience, making it easier to 
remember either the formula or the definition. 

Art. 19. Constants of Material. 

These three equations make use of the following con- 
stants of materials: 

Elastic Limit, 

Ultimate Strength, 

Modulus of Elasticity, and 

Modulus of Elastic Resilience, 
all of which have to be determined by experimental work 
on bars of the various materials. As this work cannot be 
done for each problem, a table containing average values 
of these constants for the more common materials will be 
found in the Appendix, and the question of the units in 
which each is expressed becomes important. 



14 MECHANICS OF MATERIALS 

Art. 20. Units. 

In American practice, the linear unit is the inch; the 
square unit, the square inch, and the unit of weight, the 
pound. 

In the tables, the values of E. L., U., and E., are given 
in pounds / square inch and k in inch pounds. Therefore 
when the solution of any problem requires the use of any 
of these constants, all of the quantities involving weights 
or loads must be in pounds, and those involving linear 
or square measure, in inches. 

Art. 21. Working Stresses ; Factors of Safety. 

No material is entirely free from flaws and imperfections, 
which tend to diminish the area that is effective in resist- 
ing the external force, and in no case should the stress in 
any member be greater than the elastic limit, as such a 
stress would cause some permanent deformation. Sud- 
denly applied loads, shocks, and loads producing alternate 

tension and compression, all produce stresses that are 

p 

Sfreater than the value of S = — , which is the value of S 
^ A 

for the same loads gradually and steadily applied. When 

any of these conditions occur, they tend to reduce the 

allowable or safe value of S to be used in the equation 

P = AS. Therefore, if U is the ultimate strength of the 

material, the allowable value of the unit stress can be found 

by dividing the ultimate strength by some number. Calling 

this number the Factor of Safety^ F^ S = —\^ the value of 

F 

S to be used in connection with the formula P = AS, and 

is termed the Safe or Working Stress. 

The factor F depends — 



APPLIED MECHANICS 15 

1. On the reliability of the material ; that is, the lia- 
bility of flaws or imperfections that may reduce the effect- 
ive area of the section. 

2. On the way in which the loads are applied. 

The first part has been termed a factor of ignorance, 
while the second may be determined more or less accurately 
from theoretical considerations. 

The factors of safety as given in the tables in the Appen- 
dix are those to be used in the solution of the problems 
given in this book, and it must be remembered that the 
values are only approximate. 

The safety of any structure calling for a large factor, 
while the consideration of cost always demands the 
smallest one, the final choice of the factor of safety to be 
used in any given case must be largely a question of 
engineering judgment. In some cases, as in buildings, 
the allowable stress under the various kinds of loadings 
is a part of the building laws, and the engineer has to 
conform to the law. 

For steady loads and reliable material the smallest fac- 
tor in general use is about four. 

On account of the danger of permanent injury to the 
material, no stress should exceed the elastic limit ; hence, it 
would seem better engineering to base the factor of safety 
on the elastic limit rather than on the ultimate strength, 
but such practice is not general in engineering work. 

Akt. 22. Accuracy of Calculations. 

As the use of a factor of safety of four will result in an 
area of section about twice what it would have been had 
the allowable stress been equal to the elastic limit, and the 
values of E. L., U., and E., being determined by experi- 



16 MECHANICS OF MATERIALS 

ment, are liable to an error of from 3 to 5 %, there is no 
necessity for absolute accuracy in the calculations for the 
design of the different parts of a structure. 

Students are urged to use a slide rule in the numerical 
solution of the problems given in this text, not only to 
save time during the college course, but because they will 
find the use of a slide rule almost necessary in the prac- 
tice of their profession. The slide rule should always 
give results that are not in error more than 1, or at the 
outside 2 %, which is close enough for the greater part of 
engineering calculations. 

They are warned that it is the significant figures in a 
number that is to be used as a factor, and not the decimal 
point, that is of importance. If in one case the area was 
given as 12,500 sq. in. and as .00125 sq. in. in another, 
the figure 5 is of equal importance in each case. The 
use of the latter area as .0012 sq. in. will result in an error 
of 4 % in the value of the stress. 

EXAMINATION QUESTIONS 

1. There is no relative motion between the different 
parts of a bridge, therefore each part must be in equi- 
librium. How is it that the same laws may be applied 
to machine parts, which we know have relative motion ? 

2. When may forces be considered as Axial ? 

3. Define the term "Bar" as used in the text. 

4. What is a stress ? If there are stresses in every 
section of a bar, why is it that there is no relative motion 
between the different parts ? 

5. When may stresses be termed Tensile or Compres- 
sive Stresses ? 

6. What is a Unit Stress ? 



APPLIED MECHANICS 17 

7. If a member of any structure varies in size at dif- 
ferent parts of its length, how can you find the maximum 
tensile or compressive unit stress due to an axial load ? 

8. Give some examples of forces that produce tensile, 
compressive, and shearing stresses. 

9. What are the external forces for any member of 
a structure ? What are the internal forces ? 

10. Show that P = AS^ give the units involved, and the 
limits of use for the formula. 

11. The science of mechanics is based on the action of 
forces on rigid bodies. Why is it that the same laws may 
be applied to forces acting on elastic bodies ? 

12. What is meant by the expression. Unit Deformation ? 

13. If the modulus of elasticity for steel is 30,000,000 
and for wrought iron is 25,000,000, and one bar of each is 
the same size and carries the same tensile load, which bar 
will stretch the most? 

14. The value of E may be termed a measure of the 
rigidity of a material. Why ? 

15. Define Elastic Limit and Ultimate Strength. 

16. Why does the Commercial Elastic Limit or Yield 
Point differ from the true elastic limit ? 

17. What is a Stress-strain diagram ? 

18. If a load corresponding to the ultimate strength of 
the material is placed on a bar, why is not that load 
called the Breaking Load ? Article 14 says otherwise. 
Explain. 

19. What is meant by the term Ductility ? 

20. Define Elastic Resilience. 

21. State the formulas for calculating the elastic resili- 
ence and modulus of elasticity; give the units involved and 
the limits of use for each formula. 



18 MECHANICS OF MATERIALS 

22. What is a factor of safety ? A working stress ? 
What is the difference between a working stress and a safe 
stress ? What do you understand by a safe load ? 

23. Show why, if the calculations for the stresses in 
any member are not in error more than 2^, they are 
substantially correct. 

PROBLEMS 

1. A square steel bar 2x2 in. in section and 4 ft. 
long, carries a tensile load of 60,000 lb. Required the 
unit tensile stress. 

Solution. The relation between the load, area, and unit stress, 
when the load is axial, is always given hj P = AS, hence substituting 
for F and A, from the data given in the problem, 

60,000 = i X S, or S = 15,000 Ib./sq. in. 

2. A round wooden column, 16 in. in diameter and 
12 ft. 6 in. long, supports a load of 20 tons. Required 
the unit stress. 

3. What is the value of the maximum tensile load the 
bar in problem 1 will carry ? 

4. A wrought iron bar is 2 in. in diameter and 5 ft. 
long. What tensile load may be carried if the unit stress 
does not exceed 10,000 Ib./sq. in. ? 

5. What is the maximum tensile load the bar in prob- 
lem 4 will support ? 

6. A square cast iron column is hollow, 10 x 10 in. 
on the outside and 8 x 8 in. on the inside. Required the 
maximum compressive load that may be carried. 

7. In problem 6, keeping the outside dimensions the 
same, required the inside dimensions if the load is 360,000 
lb. and the unit stress is 10,000 Ib./sq. in. 

8. A punch is 1 in. in diameter. Required the prob- 
able pressure necessary to force the punch through a 
steel plate, ^ in. thick. 



APPLIED MECHANICS 19 

Solution. In order to force the punch through the plate, the 
unit shearing stress on an area equal to the cylindrical surface 
of the punched hole must be the ultimate shearing stress of the 
material ; hence, 

P = 7rdt X S = Trx 1 X I X 50,000 = 8000 lb. approximately. 

9. Using a punch 1 in. in diameter and an available 
force of 8000 lb., what is the thickest wrought iron that 
can be punched ? 

10. An iron casting is bolted to the floor by four 
wrought iron bolts, and a force tends to slide the casting 
along the floor. Neglecting friction, what is the probable 
magnitude of the force when the unit shearing stress in 
the bolts is 10,000 lb. / sq. in. ? 

11. If the force in problem 10 was 24,000 lb., select 
four standard bolts, so that the unit shearing stress will 
not exceed 10,000 lb. / sq. in. 

12. If steel costs five cents per pound and wrought 
iron four cents, which will it be cheaper to use to carry a 
tensile load if the same factor of safety is used in each 
case ? 

Note. Assume the weights per cubic foot are the same for each ; 
then the weights in each case will be proportional to the areas of the 
sections. 

13. With wrought iron at four cents per pound, and 
other conditions the same as in 12, how much can you 
afford to pay for steel ? 

14. Required the probable elongation of the bar in 
problem 1. 

Solution. The relation between the elongation and an axial load 

PI - 

is given by ^ = — , and substituting the data as given in the prob- 
Ae 

lem, 30,000,000 = ^O^QQQ x ^^ . ... e = .024 in. 

4e 



20 MECHANICS OF MATERIALS 

15. How much work is clone by the force in prob- 
lem l ? 

16. How much work is done by the force in prob- 
lem 4 ? 

17. A concrete pier 3 ft. by 4 ft. in area carries a load 
of 300 tons. Required the unit stress. 

18. A brick pier carries the same load with a unit 
stress of 18 tons / sq. ft. Required the area of the sec- 
tion. 

19. The thickness of the head of a standard bolt is 
approximately equal to the diameter of the bolt. Com- 
pare the unit tensile stress in the bolt with the unit shear- 
ing stress in the head. 

20. A standard steel bolt IJ in. in diameter supports 

a tensile load of 9800 lb. Required the factors of safety 

for the tensile and shearing stresses. 

(The least area to resist tension is at the root of the thread. See 
tables.) 

21. If the bar in problem 1 was 4 ft. 2 in. long and 
the elongation was .025 in., required the modulus of 
elasticity. 

22. If the modulus of elasticity of wood is 1,500,000, 
required the shortening of the column in problem 2. 

23. A steel bar 1 in. in diameter has two punch 
marks 8 in. apart marked on it. The bar is placed in 
a testing machine and it is found that there is a rapid 
change in the rate of elongation, when the load was 
24,000 lb. and after a load of 47,000 lb., no more load 
could be added, the bar finally breaking between the punch 
marks, when the load was 42,000 lb. The broken pieces 
were placed end to end, and the distance between the punch 
marks was found to be 10.4 in. Required the elastic limit, 
ultimate strength, and the ductility of the material. 



APPLIED MECHANICS 21 

24. If the bar in problem 1 has the load increased 
from 60,000 to 120,000 lb., how much more work is done ? 

25. How much should the bar in problem 24 stretch 
while the additional load is being added ? 

26. A certain grade of piano wire has an elastic limit 
of 100 tons / sq. in. If the diameter of a piece of the 
wire is .05 in. in diameter, required the diameter of a 
wrought iron wire to carry the same load when the unit 
stress is equal to the elastic limit in each case. 

27. Show that the work done by an axial force on 

any bar is jK'= - — x volume, provided the elastic limit is 

2 E 
not passed. 

28. If the load in problem l is suddenly applied, 
what will be the value of the maximum stress induced 
in the bar? 

29. If a square wrought iron bar is to sustain a sud- 
denly applied load of 60,000 lb. and the stress is not to 
exceed 15,000 lb. / sq. in., required the dimensions of the 
bar. 

30. A round steel rod is to carry a tensile load of 
37,700 lb. with a factor of safety of five; required the 
diameter of the bar. 

31. Find the factor of safety in problem 1. 

32. A steel rod 2 in. in diameter in a bridge truss 
has a unit stress due to the weight of the bridge of 
4000 lb. / sq. in. A heavily loaded truck, if placed on 
the bridge, will add 51,000 lb. load to that already on 
the rod. Is it safe for the truck to cross ? 



CHAPTER TI 
APPLICATIONS 

Article 23. Bars of Uniform Strength. 

In the previous chapter, the bar was one of uniform 
section and no account was taken of its weight. When 
the bar was short, the effect of the weight of the bar 
could be neglected in comparison with the applied loads, 
and the unit stress found was that due to the loads alone. 
If the bar under axial forces is very long, the stress due 
to its own weight becomes too large to be neglected, and 
the stress in the bar is that due to the loads plus that due 
to its own weight. 

Take the case of a wire rope used to hoist a bucket from 
a deep mine shaft. The weight of the bucket and its 
contents produces a certain unit stress in the rope that is 
equal at all sections of the rope. 

If P is the weight of the bucket and its contents, and 

P 

A the area of the section of the rope, this stress is S = —, 

A. 

The sectional area of the rope at any point has to support 

the weight of rope below that point, as well as the weight 
of the bucket and its contents; therefore the section of 
the rope at the upper end being A^ and TFthe weight of 

the rope, S = — —^ instead of — -. 

A A 

It is readily seen that if W is small, the value of P + TF 

22 



APPLICATIONS 23 

is not sensibly greater than the value of P ; the value of S 

P 

will not be materially changed from -— . 

■A. 

In the case where P + TF is much greater than P, and 
the ro23e is the same size throughout, it must be large 
enough to carry a load of P + TF. 

When a long vertical bar of uniform section is under an 
axial load, it follows that every section, except one, is 
larger than necessary, and if the section area is varied so 
that the unit stress in each section is the same, the weight 
of .the bar could be reduced. Such a bar is called a Bar 
of Uniform Strength. This does not mean that the 
strength of the bar is the same at all sections, but that 
the change of section area makes the unit stress the same 
at all sections, and might better be termed a bar of uni- 
form stress, or a uniformly safe bar. 

Consider such a vertical bar, length ?, and an axial load 

P 

P. The smallest possible area, A^^ is given by A= -^i 

where S is the allowable unit stress, and design the bar 
so that S shall be constant. In Fig. 23, let A^ be the 
area at the end where the load is applied, and A be the 
area at any distance y from that section. Then at a dis- 
tance y + dy the area must be increased to A-\- clA. Let 
w equal the weight of a cubic unit of the material ; 
then the additional weight to be carried on the area 
A + dA is Aw dy^ since the term containing dA dy 
can be neglected in comparison to the term containing 
only dy. Since S is constant, and this weight is to be 
carried on the area dA^ 

J 4 Aw dy J S dA ^i ^ 

dA = — ^, OYdy :=-—-, (1) 

o IV A 



24 



MECHANICS OF MATERIALS 



gives the relation between the increase of length and the 
increase of area. If (1) is integrated, 



y = - log, A+ C, 
w 



(2) 



S 



and since A = A^, when y =0, C= -^\og A.. 

IV 

Substituting this value for C in (2) and transposing, 
log, ^ = - y 4-log, A, or logi, ^ = 0.434 f y +logio A (3) 
is an expression for the relation between the least area and 





Fig. 23. 

the area at any distance y from that section. In the 
application of the formula to a given case, different values 
might be assigned to y, and the corresponding values of A 
found, enough values being calculated to enable the pro- 
file to be drawn. In this case the outline of a vertical 
section is slightly curved. If the vertical section is made 
trapezoidal, and the bar is a masonry pier, the top of 



APPLICATIONS 



25 



the trapezoid is made proportional to Aq, and the base to 
the value of A in equation (3), when 7/ is the height 
of the pier. 

Such a pier would require more material than one of 
uniform strength ; but, although the unit stress would be 
only approximately equal at all sections, it would repre- 
sent common practice. 

Art. 24. Thin Pipes, Cylinders, and Spheres. 

Take a pipe, internal diameter D, thickness of the pipe 
wall t, carrying a water pressure of M lb. / sq. in., to find 




Fig. 24 a. 



the unit stress in the walls of the pipe. As each unit of 
length of the pipe is under the same forces, we may take 
the length as unity. Suppose a length of pipe equal to 
unity (Fig. 24: a) to be cut by a diametral plane X-X; 
then the stresses acting on the pipe walls at the section 
cut by the diametral plane must resist the pressure of the 
water tending to force the two halves of the pipe apart, 
and if ^ is small, the stress may be considered as being 
uniformly distributed over the sections of the pipe walls 
cut by the diametral plane. Therefore, if we can calcu- 
late the value of P, the force tending to separate the two 
halves of the pipe, the formula P = AS, where A is the 



26 



MECHANICS OF MATERIALS 



area of the section of the pipe walls, will give the required 
unit stress. 

A principle of hydraulics states that the pressure of 
water is the same in all directions and normal to the sur- 
face. Let Fig. 24 h represent a half section, perpendic- 
ular to the axis of the pipe. If is any angle, then 



si 

CO 

"5 



^^ 



^ 




^RrdeCosd 



Fig. 24 6. 

for a length of pipe equal to unity, the radius of the pipe 
being r, an area of the internal surface of the pipe, equal 
to rdO^ carries a pressure of M, lb. / sq. in., or the total 
radial force on that area is RrdO. This force may be 
resolved into components, parallel and perpendicular to 
the line X-X^ which is the trace of the diametral cuttino* 
plane. The components are Rr cos 6d6 and Rr sin 6d6. 
The sum of the Rr cos 6d6 forces for one half of the pipe 
is zero, and the sum of the Rr sin 6dd forces for the same 
half is 2 Rr^ or RD^ which is the force per unit of length 
perpendicular to the cutting plane, resulting from the 
internal pressure R. 

Substituting this value in the general formula P = AS, 



APPLICATIONS 27 

and noting that the area of the section of the pipe walls is 
2^, we have ED = 2 St (1) 

as a general expression for the unit stress induced in a 
longitudinal section of a pipe whose walls are thin. If 
the ends of the pipe are closed, the internal pressure of 
the water on the ends of the pipe tends to rupture the 
pipe in a plane perpendicular to the axis. 

The force acting on the ends of the pipe is evidently 

7)2 7? 

— j — , and the area to resist this force is ttDI ; hence a 
-± 

substitution of these values in P = AS gives HD = 4 St, 
showing that the stress in a plane perpendicular to 
the axis is only one half that on a plane through the 
axis. 

For a sphere with thin walls, the water pressure tends 
to produce rupture on the line of a great circle. It is 
readily seen that the pressures and areas are the same 
as for a plane section perpendicular to the axis of a cylin- 
der ; hence the same relation holds true. 

Akt. 25. Thick Pipes. 

If t is large, the stress in a plane through the axis is 
no longer uniformly distributed over the area of the sec- 
tion, but is greater on the internal radius. 

Many formulas have been proposed for finding the 
maximum unit stress in this case, the one given here 
being due to Barlow. The results are in simple form, 
and the value of the maximum unit stress being greater 
than that given by the more exact discussions, places the 
error on the side of safety. 

Barlow's formula assumes that when the fluid pressure 
acts on the internal surface of the pipe, while the diameter 



28 



MECHANICS OF MATERIALS 



is increased, the volume of the pipe walls for a unit of 
length remains unchanged. 

If we let D be the internal, and D^ be the external, 
diameters of a pipe whose walls are thick (Fig. 25), and 




Fig. 25. 



the thickness of the pipe walls ^, before the pressure is 
applied the volume of a ring one unit in length is 



7rZ>2 



(1) 



Let e and e-^ be the extensions of the diameters due to 
the fluid pressure and the volume becomes 



f(A + «iy-f(^ + «)'. 



(2) 



Expanding (2), neglecting the e^, as e is a very small 
quantity, and equating (1) and (2), the equation reduces 

*o B^e^ = Be. (3) 



APPLICATIONS 29 

The unit elongations are ■— and -J-, since the change in 

the circumference of the thin shells of diameters D and 
D^ are Tre and Tre^ while the original circumferences are 
irD and irD^ The unit stresses in the thin shells of the 
diameters D and D^, being S and aS'j, as the unit stresses 
are proportional to the unit deformations within the elas- 
tic limit, 

(4) 









e 








s 


'D 


B^e 






Si' 


^1 
^1 


'^' 


-Z>i«i 


= De, 


hence, 
e 







and substituting this value of — in (4) gives 

H 

S _Dl_rl ... 

or the unit stresses are inversely proportional to the 
squares of the diameters or radii. 

Let Sj. be the unit stress at a radius x. Then, from (5), 

and the total force exerted over the area dx times 1 is 

SJx=:Sr^^- (6) 

The integral of the left hand member of (6) is the sum- 
mation of all the stresses on one side of the pipe, and is 



30 MECHANICS OF MATERIALS 

equal to one half of the total fluid pressure tending to 
rupture the pipe ; hence the total force is 

Equation (7) reduces to RD^ = 2 St instead of RD = 2 St 
for thin pipes. Formula (7) is the one to use in all 
cases where the value of RD^ is enough larger than RD 
to cause serious error. The error in Barlow's formula 
increases as the internal radius decreases, and for thick 
pipes where the diameter is small, the more exact formulas 
of Lami and others should be used. (See Merriman's 
'^ Treatise on the Mechanics of Materials.") 

Art. 26. Riveted Joints. 

In the determination of simple stress, such as tension, 

P 

compression and shear, the formula S= — always gives 

A. 

the relation between the force P and the unit stress S. 

The area A must always be the area over which the stress 
is induced, and will, for tensile or compressive stresses, be 
a section of the bar perpendicular to the line of action of 
the force P, and parallel to the same line for shearing 
stresses. If the area of the section of the bar varies for 
different cutting planes, the plane that gives the least area 
should always be chosen. 

When two plates are joined together by means of rivets, 
the joint is called a Riveted Joint. 

Art. 27. Tension in the Plates. 

Let A and B (Fig. 27 a) be two plates joined together 
by means of rivets passing through the cover plates a and b. 
Let P be the tensile force tending to separate the plates 



APPLICATIONS 



31 



^< 



B 



a 



( 



) 



) 



A 



V 




A and B, w the width 
of tlie plates, t the 
thickness of the plates 
A and B^ t^ the thick- 
ness of the plates a and 
h^ and d the diameters 
of the rivets. 

Since P is a tensile 
force, and the greatest 
number of rivets in 
line is two, if we pass 
a plane perpendicular 
to the line of action of 
the force P through 
the line of the two 
rivets, cutting either 
of the plates A or B^ 
or the cover plates a and 5, the stress which acts in such a 
plane to resist separation is the product of the unit stress 
induced and the area cut. 

The area cut by the plane is either the thickness of the 
two cover plates a and h times the width of the plates less 
the diameters of the rivets in line, or the thickness of the 
plates A ov B into the same quantity. Hence the relation 
between the tensile force P and the unit stress induced 
in the plates is, 2 t^ (iv — 2d)Si = P, or t{tu — 2 d) Sf = P, 
depending on whether the failure is in the plates A or B 
or the cover plates a and h. 

As there is no reason why one of these sections should 
be stronger than the other, 2 ti is generally made equal to 
t. Since any other cutting plane would cut a larger area, 
the value of S as given in the above formula is the largest 



Fig. 27 a. 



32 



MECHANICS OF MATERIALS 



value possible with the force P. Figure 27 h shows the 
failure by tension in the plates. 



B 




rcy=^yn 



V 



Fig. 27 6. 




Art. 28. Shear on the Rivets. 

If the plates do not yield in tension, in order to pull 
the plate A away from B^ and the cover plates, the num- 
ber of rivets passing through A must be sheared off in 
two sections parallel to the line of action of the force P,. 
and perpendicular to the axis of the rivets. Therefore 
the area to resist shear on the rivets caused by the force 
P must be twice the sectional area of each rivet times the 
number of rivets passing through the plate A. These 

values substituted in the general formula, P = AS give 

2 7r(^2 



for this joint, 2 x 



■Ss = Ps' 



(See Fig. 28.) 



APPLICATIONS 



33 



Art. 29. Compression on the Rivets or Plates. 

Suppose that the plates resisted the ten- 
sile stress, and that the rivets, the shearing 
stress, the force P acting on J., causes the 
plates to bear on the cylindrical surface of 
the rivets. The exact effective area of each 
rivet, or the plate through which it passes, 
is not known ; but it is assumed to be the 
projected area of the rivet; that is, the 





diameter of the rivet 

times the thickness of 

the plate through 

which it passes. On 

this assumption, taking 

2 ^j = ^, the area to 

resist compression on 

each rivet is dt. This 

area times the number 

of rivets passing through A^ when 

substituted in the general formula, 

P = AS^ gives for this joint, 

2 dtS, = P„ 

as the relation between the force P 
and the unit compressive stress on 
the rivets or plates. (See Fig. 29.) 
As there is no other way that the 
joint can fail, the equation that gives the least value of P 
determines the way in which the joint is most liable to fail. 

Art. 30. General Case of Riveted Joint. 

In general, while the joint may be very long, the rivets 
are regularly spaced. In this case, the distance between 



34 MECHANICS OF MATERIALS 

the centers of any two rivets in line is called the " pitch " 
of the rivets, and P is taken as that proportion of the load 
on the entire joint that the pitch is of the length; or in 
other words, P is the load or force on the joint for a 
distance equal to the pitch of the rivets. 

Tension in the Plates. 

Taking P as above, and letting p be the pitch of the 
rivets, the relation between the tensile unit stress in 
the solid plate and the force P is, 

tpS,=^P, (1) 

where, if S is the safe unit stress, P is the safe load. 

For all joints in tension, since there can never be but 
one rivet in line in the distance p^ 

tCp-d)S, = P,. (2) 

If, as before, S^ is the safe tensile unit stress, P^ is the safe 
load when failure is considered as taking place by tension 
in the punched plates. If the values of aS'^ are the same 
in (1) and (2), it is easily seen that P^ can never equal P, 

Shear on the Rivets. 

The relation between the load P and the unit shearing 
stresses will depend on the nature of the joint. In any 
given case, the product of the number of times each rivet 
may shear, the number of rivets in the distance jt?, and the 
area of the section of the rivet perpendicular to its axis, 
will be the area over which the shearing stresses act. 
Letting c be the number of rivets times the number of 
.sections in the distance j9, then 

|ccP.9, = P.. (3) 



APPLICATIONS 35 

When safety of the joint against failure by the shearing 
of the rivets is considered, if aS'^ is the safe unit shearing 
stress, then Pg is the safe load. 

Considering (2) and (3), the values of p and d can be 
chosen so that when safe values of the unit stresses aS'^ and 
jSs are used, P^ = P^, but they are not necessarily equal. 

Compression 07i the Rivets or Plates. 

Taking aS^^ as the safe unit stress in compression, and 
td times the number of rivets in the distance p as the area 
resisting compression, and letting c^ be the number of 

"•i^'^t"' e,tdS, = P, (4) 

is the relation between the unit stress in compression and 
the load P^. As before, P^ may have different values 
from either Pf or P^, but if they are assumed to be equal, 
and safe values of aS^, aS'^, and aS^^ are used in equations (2), 
(3), (4), as there are three equations and three variable 
quantities, J9, ^, and d can always be determined. 

If the values of p, ^, and d are found in this manner, 
the joint will be equally safe against failure in all ways. 
In general, the equation which gives the least value of P 
will show the way in which failure is most liable to occur. 

Art. 31. Kinds of Riveted Joints. 

Lap Joints. Here the two plates to be joined together 
lap by each other and the rivets pass through both plates. 
The rivets tend to shear on but one section, and are said 
to be in "single shear." 

Butt Joints with Single Cover Plates. Here the plates 
are both in the same plane, and the joint between them is 
covered by a plate of the same thickness as the plates. 
Any rivet passes through the cover plate and one of the 



36 



MECHANICS OF MATERIALS 





Double riveted lap joint. 



Double riveted butt joint with 
single cover plate. 



1 




i 



1^ 



Triple riveted lap joint. 




Double riveted butt joint with double cover plates. 
Fig. 31. 



APPLICATIOXS 37 

plates that are to be joined together. The conditions for 
shear and compression are evidently the same as for lap 
joints. 

Butt Joints with Double Cover Plates. In this kind of 
a joint the plates are in the same plane, and the cover 
plates, each one half the thickness of the plates to be 
joined, are placed on either side of the joint. Any rivet 
passes through both cover plates and one of the plates that 
are to be joined. 

An inspection of the figure will show that each rivet is 
liable to be sheared in two sections and is said to be in 
" double shear," while the conditions for compression are 
the same as for the one with single cover plates. 

Either type of a joint may have one or more rows of 
rivets, and the pitch in all rows is generally the same. 
The joint is said to be Single^ Double^ or Triple riveted, as 
there are one, two, or three rows of rivets. The figures 
show the details of the various joints and styles of riveting. 
It is evident that if lines are drawn passing through any 
two adjacent rivets in the same row, and parallel to the 
line of action of P, the rivets included between these lines 
will be the number of rivets that are to be considered as 
resisting the shear and compression. 

Art. 32. Efficiency of a Riveted Joint. 

When P;, P5, and P^ are the maximum safe loads a 
riveted joint will carry, the efficiency of that joint may be 
■defined as the ratio of the least of the above values, to the 
load the unpunched plate of the same length will carry 
under the same conditions. 

From this definition it is evident that the efficiency of 
any joint is P^, P^, or P^, divided by P, depending on the 



38 MECHANICS OF MATERIALS 

relative values of P^, P^, and P^. Of all the ways in 
which a riveted joint may fail, the failure by compression 
of the rivets or plates is the least understood, and many 
engineers design the joint for equal strength in tension 
and shear, and simply check the resulting dimensions for 
the compressive stress. This practice has resulted in the 

efficiency of a riveted joint being given as •, but this 

P 
is only approximately true unless P^= P^ = P^. 

In general, when the values of t and d are calculated, 
the nearest commercial sizes have to be chosen, and the 
values of P^, P^, and P^ are rarely ever equal. 

In many cases the pitch is fixed by the conditions for 
the tightness of the joint against leakage, as for boilers, 
tanks, and pipes, and in such cases only two conditions 
can be satisfied. 

In the development of the preceding formula no account 
has been taken of the friction that must exist between the 
plates through which the rivets must pass. 

As there is no good theoretical way of introducing the 
resistances due to friction in the formulas for strength, 
riveted joints have been pulled apart in testing machines, 
and the accuracy of the formulas checked by the breaking 
load as determined by the test. While the results in 
many cases seem to show that the theory that has been 
given here does not hold true, the conditions that are con- 
current with the rupturing load not being the same as when 
all the stresses are within the elastic limit, there seems to 
be no good reason why the formulas as developed will not 
give reliable results. 

In the design of a riveted joint for a pipe or a boiler to 
carry a pressure of R lb. /sq. in., as one half of the 



, APPLICATIONS 39 

internal pressure tending to disrupt the pipe or boiler 
is carried on one joint of the shell the value of P to be 
used in the formulas is one half of the total pressure 

acting over a length p^ or, — —^ = P. 

Art. 33. Stresses Due to Change of Temperature. 

All metals tend to change in length as their temperature 
changes. If the change is resisted, that resistance must 
cause a stress in the material. 

Consider a bar I units in length, A units in area, free 
to change its length as the temperature changes. If the 
change in length due to a given change of temperature is 
g, and a force is exerted to restore the bar to its original 
length, the unit stress induced by that force will be given 

e 
Therefore if a force prevents the change from taking 
place, it must induce an equal unit stress, and this unit 
stress is independent of the area of the bar. Knowing 
the change in unit of length for a change of 1° of tem- 
perature, or the coefficient of linear expansion, the unit 
stress in any bar corresponding to any change of tem- 
perature may be found provided the unit stress is within 
the elastic limit of the material. 

If the bar is under an initial unit stress before the 
change of temperature, the change will increase or de- 
crease that stress, depending on the nature of the initial 
and temperature stresses. 



40 MECHANICS OF MATERIALS 

PROBLEMS 

1. How long will a bar of wood have to be, in order 

that its own weight will produce a unit stress of 300 lb./ 

sq. in. ? The bar is hanging vertically. 

Solution. Taking the weight of a bar of wood, 1 sq. in. in sec- 
tion and 3 ft. long, as ^§ lb., the bar will have to be as long as 300 
divided by ^f, equal 360 yd. Ans. 

2. What is the length of a vertical steel bar a square 
inch in area, that carries a tensile load of 40,000 lb., at 
the lower end when the maximum unit stress is 15,000 
Ib./sq. in. ? 

3. Find the probable elongation in problem l. 

Solution. Since the maximum unit stress is 300 lb. / sq. in. and 
the minimum 0, the average unit stress must be 150 lb. / sq. in., 

and we have given that E = '^-, 

e 

1,500,000 = 150 X 360 x 36^ .^ ^ ^ ^296 in. 
e 

4. Find the total elongation in problem 2. 

(Total elongation is that due to the load and its own weight.) 

5. Find the height of a brick chimney of uniform 
section, when the maximum compressive unit stress is 18 
tons /sq. ft. 

6. Suppose that the sectional area of the base of a 
chimney was twice that at the top, and that the change in 
area was uniform, how high could the chimney be built if 
the limiting value of the unit stress was 18 tons /sq. ft. ? 

7. Find the areas of the top and bottom section of a 
stone pier, 100 ft. high, to carry a load of 240 tons, the 
unit stress in all sections to be constant. 

8. If the pier in problem 7 had the top and bottom 
areas as found and the vertical section was trapezoidal, 
find the unit stress at the bottom of the pier. 



APPLICATIONS 41 

9. The wire rope used for hoisting in a certain mine is 
1^ in. in diameter, and weighs 2.5 lb. /ft. If the mine 
is 800 ft. deep and the safe working load for the rope is 
5^ tons, what weight may be raised ? 

10. A pipe 6 in. in diameter is to carry water under a 
pressure of 1000 Ib./sq. in., with a factor of safety of 
6; required the thickness of the pipe w^alls. 

11. A standard 2-inch steel pipe is 2.375 in. outside 
diameter and 2.067 inside. This size is tested under a 
pressure of 500 Ib./sq. in.; required the unit stress in 
the pipe walls. 

12. A steel pipe 10 in. in diameter is to carry water 
under 2770 ft. head. The factor of safety is to be 10. 
Find the thickness of the pipe. 

(A column of water 1 ft. high and 1 sq. in. in area weighs .431 lb.) 

13. Check the results in problem 12 by Barlow's for- 
mula and find the unit stress. 

14. Compare the maximum unit stress in the pipe of 
problem 11, as determined by the formulas for thick and 
thin pipes. 

15. Write the formulas for determining the strength of 
the following riveted joints in tension, compression, and 
shear. The pitch is p^ the thickness of the plates f, diam- 
eter of the rivets d, and the safe unit stresses in tension, 
compression, and shear are Sf, >S'g, and S^. 

(a) Single riveted lap joint. 

(5) Double riveted lap joint. 

(c?) Single riveted butt joint with one cover plate. 

(d) Double riveted butt joint with one cover plate. 

(e) Single riveted butt joint with two cover plates. 
(/) Double riveted butt joint with two cover plates. 
(^) Triple riveted butt joint with two cover plates. 



42 MECHANICS OF MATERIALS 

Solution for (a). 

If Pt is a tensile force acting on the joint for a distance equal to 
the pitch, then since P = AS, and the area to resist the tensile 
stress is t{p — d), Pt = t(p — d)St gives the relation between the 
load and the unit tensile stress. 

Let Pc be the tensile force that brings compression on the rivets. 
As there is but one rivet in the distance j>, the area to resist compres- 
sion is id, and since P = AS, Pc = tdSc is the relation between the 
load and the compressive unit stress, and if Pg is the tensile force that 
produces shear on the rivets, as there is only one rivet in the distance 

p, and it can shear in but one section, SiS P = AS, Pg = Sg is the 

relation between the load and the unit stress in shear. 

16. If the values of P^, P^^ and P^, in each of the dif- 
ferent joints given in problem 15 are taken as being equal, 
find the expressions for the values of p and c?, in terms of 
t, and the efficiency of each joint. 

Solution for (a). ,o _7TfP ^ 

ttSs 
and t(p -d)St = tdSc, 

^ d(S, + S,) 

^ St 

Substituting the value of d, 

^ ^ 4 >Se t(Sc + St) ^ 

TT Sg St 

and since the expression for the efficiency when the strength of the 
joint is equal against all kinds of stress is ^ ~ — -, substituting for 

these their values as found, their efficiency is 



Sc -}- St 

17. A steam boiler, 60 in. in diameter, carrying 120 
Ib./sq. in., is to have the longitudinal seams double riv- 
eted butt joints with two cover plates. Take S^ = 12,000 
Ib./sq. in., aS'^ = 10,000 Ib./sq. in., and make the joint 
equally safe against failure by either tension or shear. If 



APPLICATIONS 43 

the efficiency is to be either approximately 75 %, required 
the thickness of the plates, the diameter, and pitch of the 
rivets. 

Solution. The thickness of the plate is given by RD = .75 • 2 St 
as the unit stress in the unpunched part of the plates can be only 
75 % of the allowable unit stress, or, 

120 X 60 = .75 X 2 X 12,000 t 

^ ^ 120 X 60 ^ Q ^ 
.75 X 2 X 12,000 

and the expression for the efficiency in tension being efficiency 

^^^^=.75, 
P 

p = 4:d. 

Take y'g as the nearest market size for the required thickness of the 
plate, for equal strength, 

t(p-d)St = 2 X 2 x^'^,, 

■4 

^\ (4:d - </)12,000 = 7rf/2 X 10,000, 

d = \" approximately; 
then ^ = xV', c?i=i", andj9 = 2". 

18. In problem 17, taking the values j9, ^, and d as 
found, what must be the value of the unit stress in com- 
pression, in order that the joint will be equally safe 
against failure by tension, compression, or shear ? 

19. A boiler, 30 in. in diameter, has double riveted lap 
joints, plates |- in. thick, rivets |- in. diameter, pitch of 
the rivets 2.5 in. Taking Sf as 60,000 Ib./sq. in., find 
the pressure per square inch that may be carried with a 
factor of safety of 6, considering failure by tension of the 
plates alone. 

20. What are the values of aS'^ and S^^ and the efficiency 
of the joint, in problem 19? 

21. A triple riveted butt joint with two equal cover 
plates is to have an efficiency of 80 %. Using S^ = 10,000 



44 MECHANICS OF MATERIALS 

lb. / sq. in. as the safe unit stress in tension, what will be 
the values of S^ and /S'^, when the joint is equally safe 
against failure by tension, compression, or shear ? 

22. If *S', is taken as 10,000 Ib./sq. in., and S^ as 15,000 
lb. /sq. in., what is the highest possible efficiency of a 
double riveted lap joint, designed for equal strength 
against tension, compression, and shear ? 

23. A steam boiler with double riveted lap joints is 
to carry 125 lb. /sq. in. pressure. The allowable tensile 
unit stress is 10,000 lb. / sq. in. The j^lates are f in. 
thick, rivets 1 in. diameter, and the pitch 3J in. What 
will be the largest possible diameter of the boiler ? 

24. Compare the values of aS';, S^^ and S^ in problem 

23. 

25. A 30-foot steel railroad rail undergoes a change of 
temperature of 100° . If the change in length is pre- 
vented, what unit stress will be set up in the rail ? 

Solution. The coefficient of linear expansion for steel is .0000065 
per degree ; hence the unit elongation for 100° is .00065 in., 
and since 

S = Ee, 

S = 30,000,000 X .00065 = 19,500 lb. / sq. in. 

26. For electric railway work, the steel rails are often 
welded together. Assuming that there is no change of 
length, what is the maximum range of temperature allow- 
able if the unit stress is not to be greater than the elastic 
limit? 

27. The walls of a building had bulged out, and to pull 
them into place, five steel rods each two sq. in. in area 
were passed through from one wall to the other. The 
temperature was then raised 100° and the nuts on the 
rods tightened, so that the load on each bolt was 1000 
lb. When the rods are at the original temperature, what 
is the maximum pull they could exert on the walls ? 



APPLICATIONS 45 

28. At St. Louis, Mo., a battery of steam boilers was 
connected together by a pipe in which there was no pro- 
vision made for expansion. The temperature of the 
steam was about 360° F. and that of the room, 100°. 
Assuming that there was no change in length, what was 
the maximum unit stress in the pipe due to the change of 
temperature ? 

(«) Material of the pipe steel ? 

(^) Material of the pipe cast iron ? 



CHAPTER III 



BEAMS 

Art. 34. Kinds of Beams. 

When a bar is placed in a horizontal position, and acted 
on by forces perpendicular to the axis of the bar, it is 
called a Beam. 



[ 



1 



m 

Cantilever Beam. 
Fig. 2Aa. 




Cantilever Beam. 
Fig. 346. 




■ 



w 



^fl, 



R, 



\ 



Simple Beam. 
Fig. 34 c. 



Continuous Beam. 
Fig. 34 d. 



i 



^^4 



If the beam has two supports on which it merely rests, 
it is called a Simple Beam. 

A cantilever beam has only one support, which is at the 
middle, or, what is the same thing, has one end firmly 
fixed in the wall, leaving the other end free. 

When a beam has both ends firmly fixed in the walls, 
or one end fixed and the other merely supported, it is 
called a Fixed or Restrai7ied Beam. 

A beam supported at more than two places is called a 
Continuous Beam. 

46 



BEAMS 47 

Akt. 35. Reactions at the Supports. 

The reactions at the supports are the forces acting 
between the supporting walls and the beam, and so far as 
the beam is concerned, they may be treated as vertical 
forces acting upward. 

Art. 36. Uniform and Concentrated Loads. 

The loads on a beam are the Aveights that the beam 
carries, and since the attraction of gravity always acts 
downward, they may be represented as vertical forces. 
When the load is distributed uniformly over the entire 
length of the beam so that each element of the length 
of the beam carries the same load, the load is said to be a 
Uniform Load. 

When a load is carried on so small a portion of tlie length 
of the beam that the effect of the weight acting as it does 
over that small portion may be assumed to have the same 
effect as a single force acting at the center of the load, it is 
called a Concentrated Load. 

Since a simple or cantilever beam under any loads may 
be considered as a body acted on by forces which keep it 
at rest, the laws relating to the equilibrium of forces must 
be satisfied. 

In general, all the forces, loads and reactions, will be 
vertical, and the magnitude and position of the loads will 
be known, so that the magnitude of the reactions may be 
determined by applying the laws relating to the equilibrium 
of parallel forces. 

These laws are: 

The algebraic sum of all the forces equals zero, and 
the algebraic sum of the moments of all the forces about 
any point equals zero. 



48 



MECHANICS OF MATERIALS 



These two equations are sufficient to determine the 
reactions for simple and cantilever beams, as there are not 
more than two quantities to be determined. 

For all other beams another condition is derived by the 
use of the theory of the Elastic Curve. (See Chapter IV.) 

The length of a simple beam is the distance between 
the supporting walls and the distance the beam projects 
beyond the wall for a cantilever beam. 

While in any case the beam must rest on the supporting 
wall for a finite distance, the point of application of the 
single force that is to replace the resultant of the forces 
acting between the beam and the wall is taken at the edge 
of the wall beyond wdiich the beam projects. 

Let Fig. 36 represent 



^ 



>--^-^^: 



A--I 



i 






W 
Fig. 36. 



-h- 



W9 



1 



a simple beam, length Z, 
weight Tf, carrying two 
concentrated loads Pj and 
P2 at distances p^ and jt?2 
from the right reaction, 

and the values of the reactions R^ and R2 are required. 
The weight W may be considered as a uniform load, 

and for equilibrium. 



R, + R2-Pi-P2- ■^= 0. 



(1) 



Taking moments about a point in the line of action of 
R2, and giving the moment a positive sign when it tends 
to produce clockwise rotation. 



RJ^R^O - F,p, - F2P2 - -^ = 0. 



(2) 



The term containing R2 is zero, therefore Ri may be 
found, and by substituting for R^ in equation (1), R2 may 
also be determined. 



BEAMS 



49 



R^ may also be found by taking moments about a point in 
the line of action of j^i, and this value used as a check. 

If the weight of the beam, TF5 included both the weight 
of the beam and a uniform load, the equations would have 
been the same. 

Art. 37. Vertical Shear. 

Let Fig. 37 a represent a simple beam, loaded with both 
uniform and concentrated loads, and Fig. 37 5 a cantilever 
beam with the same loading. 

Suppose either beam to be cut by a plane X-X perpen- 
dicular to the axis of the beam, at any distance x from the 



X 




-X 



Ri 



Fig. 37 a. 



R 



1 




Fig. 37 &. 



left end of the beam, and consider the end marked U. This 
end is acted on by known forces, since, when the loads are 
known, the reactions can be found, and relative to the end 
marked F, these forces tend to produce translation either 
up or down, the magnitude of the resultant force being 
the algebraic sum of all the forces acting on the part JEJ. 

Similarly, considering the part marked F, the resultant 
of all the forces acting on F tends to produce translation 
relative to F, and these two resultants being equivalent 
to all the forces acting on the beam, must be equal and 
opposite in sign, since the beam is in equilibrium. 

These two resultants are a pair of shearing forces, and 
either one is the force producing shear in the plane X—X^ 
and if we wish to call one of these forces the Vertical 



50 MECHANICS OF MATERIALS 

Shear at the section X-X^ it will be necessary to define 
vertical shear so that the sign will be determined as 
well as the magnitude. 

The vertical shear at any section of a beam is defined 
as follows : 

The Vertical Shear for any section of a beam is the 
algebraic sum of all the forces acting on that portion of 
the beam, lying to the left of that section. 

When the resultant force acts upward, the vertical 
shear is considered positive, and negative when it acts 
downward. 

Art. 38. Bending Moment. 

If we take the sum of the moments of all the forces 
that act on the right and also those that act to the left 
of the section X-X (Figs. 37 a and 37 5) about a point in 
that section, the resulting moments must be equal and 
opposite in sign, and as either is a measure of the tend- 
ency of rotation to take place about a point in the plane 
X-X^ they are the bending moments for that section. 

In order to fully determine the bending moment both 
as to sign and magnitude, it is defined as follows : 

The Bending Moment at any section of a beam is the 
algebraic sum of the moments of all the forces, acting on 
that portion of the beam lying to the left of the section, 
moments being taken about a point in that section.* 

It is considered positive when the moment tends to 
produce clockwise motion and negative for counterclock- 
wise motion. The "section of the beam" as used in the 
definition of both vertical shear and bending moment 

* A cantilever beam is always considered as being fixed at the right 
end, leaving the left end free. If the beam projects toward the right, 
look at it from the other side. 



BEAMS 51 

refers to any plane section perpendicular to the axis of 
the beam, and the " forces acting to the left of the sec- 
tion " includes both the loads and reactions whose points 
of ap23lication are on the left of the section. 

Art. 39. Resisting Shear. 

Since the part of the beam on the left of the section 
X-X (Figs. 37 a and 37 5) is acted on by external forces, 
and as a part of the whole beam it is in equilibrium, there 
must be internal forces acting in the section X—X^ which 
taken with the external forces acting to the left of the 
section, constitute a system of forces in equilibrium. 
Suppose these unknown forces to be resolved into their 
horizontal and vertical components. Then, since equi- 
librium exists, the algebraic sum of the vertical compo- 
nents of the internal forces must equal the sum of the 
vertical forces, and since the external forces have no 
horizontal components, the sum of the horizontal compo- 
nents of the internal forces must be zero, and also, tlie 
algebraic sum of the moments of the external forces must 
equal the sum of the moments of the internal forces, mo- 
ments being taken about a point in the section X-X. 

From the first condition, if we give the name of Resist- 
ing Shear to the sum of the vertical components of the 
internal forces. 

The Vertical Shear = the Resisting Shear^ 

and assuming the shearing forces to be uniformly dis- 
tributed over the area of the section, then 

r= AS, (3) 

where A is the area of the section and S is the unit shear- 
ing stress, and I^the vertical shear for the section. 



52 MECHANICS OF MATERIALS 

Art. 40. Resisting Moment. 

From the second condition, since the horizontal com- 
ponents of the internal forces must be either tensile or 
compressive forces, the sum of the tensile Forces = the 
sum of the compressive Forces. 

Giving the name of Resisting Moment to the moments 
of the internal forces about a point in the section, the 
third condition for equilibrium states that, 

The Bending Moment = the Resisting Moment. 

The relation between the bending moment of the exter- 
nal forces and the unit stresses in the section considered, 
cannot be found by the laws of mechanics alone, as the 
distribution of the internal forces is unknown. The in- 
formation necessary may be derived from the results of 
experimental observations on beams while under the 
action of bending forces. 

When a beam is under the action of bending forces, it 
is observed that along the concave surface of the beam 
the fibers * of the beam are shortened, while those on 
the convex surface are lengthened, and that along a 
certain plane section of the beam there is no change in 
length. 

We know that a compressive force shortens, and that a 
tensile force lengthens, any bar on which it acts, and that 
where there is no deformation there can be no force act- 
ing ; therefore the stress on the concave surface must be 
compression, and that on the convex surface, a tensile 

* The word "fiber" as used here may be defined as a bar of elemen- 
tary sectional area and a length equal to that of the beam, the whole 
beam being composed of a bundle of such fibers. It is not necessary 
that the beam should be of a fibrous material in order that this conception 
should be true. 



BEAMS 53 

stress, while at the certain plane, called the Neutral Sur- 
face, there is no stress of any kind. 

When the loads were such that there was no unit stress 
greater than the elastic limit, it was observed that the 
deformation of any fiber was proportional to its distance 
from the neutral surface. 

If we call the trace of this neutral plane on any plane 
section of the beam perpendicular to the axis, the Neutral 
Axis of the section, since there is no unit stress greater 
than the elastic limit, it is evident that the unit stress at 
any point in that section, and consequently the forces pro- 
ducing that unit stress, must vary directly as the distance 
from the neutral axis. This assumes that U is constant, 

since U = — , and when the location of the axis is known, 

e 

the unit stress at any point may be found. 

Let Fig. 40 represent any cross section perpendicular to 
the axis of the beam, and the line X-X the neutral axis. 
From the experimental observations we know that the 
greatest unit stress must be at the 
greatest distance from the neutral 
axis ; and letting c be the distance 
from the neutral axis to the fiber 
most distant from that axis, A the 
area of the section, dA the area of ^^' *^' 

any fiber, y the distance of that fiber from the neutral axis, 
and S the unit stress at a distance c from the neutral axis, 
then since the force varies as the distance from this neu- 
tral axis, the force at any distance «/, acting on the area of 

any elementary fiber dA is —yd A, and — I ydA is the sum- 

mation of the horizontal forces acting on the whole section. 




54 MECHANICS OF MATERIALS 

From the conditions of the problem, this sum is equal 

to zero, and as neither aS' or c can be zero, I ydA must be 
zero. 

If we assume the density to be constant, this is the con- 
dition where the axis of moments passes through the cen- 
ter of gravity of the section ; hence the neutral axis 
passes through the center of gravity of the section. The 

force on any fiber being —ydA^ the moment of this force 

c 

S 
about the neutral axis is — y'^dA^ and the sum of the mo- 

e 

ments of these forces about a point in the section is 
— I y'^^dAy which is the Resisting Moment by definition. 
The iy'^dA is defined in mechanics as the moment of in- 
ertia of the section about a gravity axis, and is repre- 
sented by the symbol I. 

Therefore, since the Bending Moment = the Resisting 

Moment, ^j 

M = —, 00 

c 

In this formula il^f is the bending moment of the exter- 
nal forces that act on the left of any section, 7" the moment 
of inertia of the section about a gravity axis perpendicu- 
lar to the direction of bending, c the greatest distance of 
any fiber from the neutral axis, while S is the maximum 
unit stress in that section. 

The formula expresses the relation between the bend- 
ing moment and the unit stress in the section, and if the 
maximum unit stress in a beam is desired, 31 must be the 
maximum bending moment for that beam under the given 
loads. 



BEAMS 3B 

As an aid to memory, attention is called to the simi- 
larity between this expression and the one derived for 
axial stress. 

P and M are the external forces, S in each case is the 

unit stress induced; and as - depends on the shape and 

c 

area of the section for its value, it may be considered as 
replacing A in the formula for axial stress. 

Art. 41. Use of Formula. 

In the derivation of the formula for axial stress, there 

was no consideration taken of the intensity of the stress 

^ or of the nature of the material, the formula holding true 

for all unit stresses and materials. When the formula 

Mc 

S = — - was derived, certain conditions were specified. 

They are : 

(1) The material was to be elastic, and since we assumed 
that the forces were proportional to the deformations, the 
modulus of elasticity must also be constant. 

(2) In order for I t/^dA to be the sum of the moments 

of the differential areas about the gravity axis, the material 
of the beam must have a uniform density. 

(3) No unit stress to be greater than the elastic limit. 
If the material and loading of a beam does not satisfy 

Tl'T" 

these three conditions, the formula S = — - will not give 
the true unit stresses. 

The Modulus of Rupture is the value of ^S' as derived 

from aS' = -— , when M is large enough to rupture the 

beam. Since the formula only holds true for unit stresses 
within the elastic limit, the value of the modulus of rup- 



5Q MECHANICS OF MATERIALS 

ture as an engineering constant is at least doubtful. In 
testing cast iron bars in bending, the breaking load at the 
center, which is of course proportional to the modulus of 
rupture, is taken as a measure of the quality of the material. 
The results of such tests are useful for comparison only 
when the tests are made on bars of the same length and 
size. 

The formula S = — — , expressing as it does the relation 

between the bending moment and the unit stress in a 
beam, is used in all calculations for the strength, safety, 
and design of beams. When sufficient data are given to 
fully determine the value of M^ the value of either S or 

— may be found. 

^ I 

The value of - depends on the form and area of the sec- 

c 

tion, and is called the Section Modulus. 

Mc 

In the application of the formula S = -— to any given 

case, the question of units becomes one of great importance. 

It does not make any difference in the effect of J/, 
whether it is expressed in inch pounds, or foot pounds; 
but as S is the unit stress and is usually given in pounds/ 
square inch, the value of M must be expressed in inch 
pounds and c and I in. inches, in order to get correct results. 

The ton could just as well be used as the unit of weight 
and the foot as a unit of length, but such practice w^ould 
require a special table of the constants of materials, the one 
given in the Appendix being based on the pound and inch. 

As there are many different sections that have the same 
section modulus, the designer is called on to choose a form 
of section best suited to the conditions that exist in the 
case in hand. 



BEAMS 



57 



The value of I for a rectangular section, breadth 5, and 

depth d^ about an axis through the center of gravity 

hd^ d 

parallel to the side 6, is -zr^i and c is -. 

Substituting these values in tlie gen- 

3Ie 

eral formula S = -— gives the value 




of i!f as 



Sbd^ 



I 



Therefore, when the section is rec- 
tangular we see that the value of M 
for any given value of S increases 
directly as b and as d^^ and any in- 
crease in the value of d increases the 
strength more than a proportional in- - 
crease in the value of ^, while the 
weight of the beam will be the same 
for either case. 



w 







-2j02 



d 



Z. 61 

WT. 29.3 TO 34.6 LBS. 



\^ 



1 



-2-4f 



"TN 



^O 



Fig. 41 a. 
A practical limit of the ratio of t is about 6. 

From the known condition 
that the unit stress in any point 
in the section varies as the dis- 
tance from the neutral axis, it 
is evident that the form of sec- 
tion which presents the great- 
est area where the unit stresses 
are large and a minimum area 
^' ^''^ where they are small will be 
the best from an economic 
Fig. 41 6. standpoint. 

The common steel I beam, so called from the form of the 
section, is an example of this distribution of area. There 



<■%-> 



Kl 



% 



-3J/,- 



58 



MECHANICS OF MATERIALS 



^T. 97 

WT. 9.3 LBSy 
3.1/2'^^ 



is always a maximum vertical shear to be resisted in all 
beams, and care should be taken that the area of the sec- 
tion chosen is large enough to resist 
-.'I'the shearing forces. In most cases if 
the beam is safe against the bending 
forces, it will be safe against the 
shearing forces, but the 
unit shearing stress 
should always be inves- 
tigated before the final 
decision is made as to 




Fig. 41c. 
the size of a beam. 

The hollow box or cored sections in cast 
iron machine parts subjected to bending 
forces represent the best practice on account 
of the large value of I relative to the weight. 

Plate girders", made 





"W 



- .^ 

01 JO 



01 



O 



z 



-1-1 
Fig. 41cZ. 



7 



by riveting angle 
JL. irons to a steel plate 
Plate Girder. Called the web, making a beam 

Fig. 41 e. whose section resembles that of the 

common I beam, are in common use in structural steel 
construction. 



Art. 42. Shear and Moment Diagrams. 

If a line which may be either straight, curved, or 
broken be drawn so that the ordinate to that line from 
any point of a straight line representing the length of the 
beam equals the vertical shear or bending moment for that 
section of the beam, the resulting diagram is called a Shear 
or Moment Diagram^ depending on whether the vertical 
shears or the bending moments are used as ordinates.* 
* This line will be called the shear or moment line. 



BEAMS 59 

To draw either diagram, the shears or moments for 
each unit of length of the beam might be calculated and 
the results plotted to scale, using as ordinates the values 
of the moments or shears and as abscissa the distances of 
the sections from the left end of the beam. A line 
through the points so located would be the shear or 
moment line, as the case might be. 

This is a tedious process and may be shortened by a 
study of the effect of the different kinds of loads on the 
form of the shear or moment line. 

Art. 43. Shear Diagram. 

For a concentrated load, the difference between the 
shears for sections taken just to the right and left of the 
point of application of the load is the magnitude of that 
load, since in the former case the " negative forces acting 
to the left of the section " are increased by the magnitude 
of the load over those for a section taken just to the left of 
the load. Therefore, the shear line will always contain 
a line perpendicular to the length of the beam under each 
concentrated load. 

When only concentrated loads are considered, the shear 
line between any two concentrated loads will be a straight 
line parallel to the length of the beam, since there is no 
change in the forces acting to the left of any section 
taken between the two loads. 

For uniform loads of w pounds per linear unit, the 
shear line will be a straight line inclined toward the 
right, as the resultant force on the beam due to the uni- 
form load is decreased by ivx^ where x is the distance of 
the section from the left end. 

If there are concentrated loads as well as uniform loads, 



60 



MECHANICS OF MATERIALS 



the shear line will be straight and vertical under the 
loads and inclined between the concentrated loads. 

To apply these principles, let Fig. 43 represent a 
simple beam, carrying two concentrated loads, P^ and P^, 
and a uniform load of w pounds per linear unit. 

Remembering the definition of vertical shear, it is 
easily noted that the shear at the left end is equal to 
the left reaction R^ which is plotted as AG. The shear 
between the left end and P^ is R^ — ivx, where x is the 
distance of the section considered from the left end. 

For a section distant j?^ from the left end, taken just 
to the left of P^, the shear is less than the shear to the 
left end by wp-^, and the ordinate to the line AB at any 

point will be the vertical shear 
for that section. For a section 
just to the right of P^ the 
equation of the shear has 
become R^ — wx — P^, as the 
two sections taken to the right 
and left of P^ are considered so 
close together that the uniform 
load has not increased. The 
shear line will therefore drop 
to (7 on a vertical line through 
the point of application of P^. 
Between P^ and P^ the shear will decrease at the same 
rate as between R^ and P^ since the load increases di- 
rectly as the distance and CB will be parallel to AB. 
Then comes the drop due to P^, and UF is parallel to AB 
and CJ). FO must be the value of the right reaction, 
since the sum of the vertical forces must be equal to zero. 
It is evident that the consideration of more concen- 




FiG. 43. 



BEAMS 61 

trated loads would simply extend the diagram in a similar 
manner, and also, if the magnitude of the uniform load per 
unit of length should change at any point, the inclination 
of the shear line would also change at the same point. 

Art. 44. Moment Diagrams. 

The general equation of the bending moment for any 
section of a beam may be written from its definition. 
For a section of a beam distant x from the left end, the 
moment of the left reaction R^ about a point in that sec- 
tion is R^x^ and the moment of the uniform load on the 
left of the section about a point in that section is the 

X • 1VX 

arm - times wx^ or — — , hence, 

2 2 

2 ftlie sura of the moments of the loads] 

3^= R^X — — < that act to the left of the section I 

"^ [about a point in that section. J 

If there are no concentrated loads, the term containinsr 
the sum of the moments, etc., is zero, and for a cantilever 
beam the term containing the reaction R^ is zero, since 
there is no left reaction. 

In some cases the supports of a beam are not at the ends, 
and in that case the moment of the left reaction would be 
R^(^x—the distance of the reaction from the left end). 
When there are uniform loads on the beam, the above 
equation shows that M varies with x^^ hence the moment 
line will be a curve. 

If there are no concentrated loads, the equation of the 
curve will be the same at all sections, being a parabola 
whose equation is at _ j? _w^ 

When there are concentrated loads in connection with 
the uniform load, the form of the equation changes at each 



62 MECHANICS OF MATERIALS 

concentrated load, and the moment line, while still j)ara- 
bolic in form, has a different equation between each pair 
of concentrated loads. 

If only concentrated loads are considered, the equation 
of the bending moment is 

{the sum of the moments of the loads acting! 
to the left of the section about a point in I 
that section. J 

If 2:j is the distance from P^ of any section of a beam 
taken between any two concentrated loads P^ and P^, the 
moment of P^ about a point in that section is PiX^ It is 
evident that the equation of the bending moment for any 
section will be changed by the addition of PiX^ the instant 
the section is taken to the right of P^ and as at that point 
x-^ is very small there will be no abrupt change in the value 
of the bending moment as the section passes under P^. As 
the value of ilf depends on the first power of x^ it is evident 
that the form of the equation is that of a straight line. 

Therefore, wdien only concentrated loads are considered 
the moment line will consist of a series of straight lines 
whose inclination changes at each concentrated load. If 
the loads are all concentrated, the bending moments may 
be calculated for sections under the loads and plotted to 
scale. Joining the points so plotted by straight lines 
will accurately determine the form of the moment line. 
If uniform loads are to be considered, the moment line be- 
tween any two loads being a parabola, the bending 
moments for enough sections between any two loads must 
be found to enable the curve to be drawn (Fig. 43). 

The value of M'\n the general equation for the bending 
moment is zero when x is zero, hence 3/ is zero at the left 
end. The moments of all the forces about any point being 



BEAMS 63 

zero, M must also be zero at the right end. That this 
is true for simple beams is so apparent that no proof is 
needed. In the case of cantilever beams with the right 
end fixed in the wall, if we remember that such a beam is 
but one half of a beam supported at the middle and that 
the right end of such a beam is, strictly speaking, the 
middle, the truth of the statement is evident. 

Therefore the moment diagram will be a closed figure 
in all cases. 

Akt. 45. The Relation between the Vertical Shear and 
the Maximum bending Moment. 

Writing the general equation of the bending moment 
M=R^x-'^-P^ix-p,)-P^{x-p^) . . .P„(x~p„-) 

where Pi P2 - - • Pn ^^^ ^^ ^^^ tlie distances of the loads 

and section from the left end of the beam and the loads 

J^j, P^^ ' • • i^„ act on the beam to the left of the section. 

The value of x which makes 31 11 maximum is the value 

that renders ~^— = or 

dx ^ 12 

The right hand member of the last equation is the ex- 
pression for the vertical shear for any section of a beam, 
therefore the value of x which makes the vertical shear zero 
renders the bending moment a maximum. 

As the equation for M was a general one and will apply 
to all kinds of beams and loadings, the results are true 
for all cases. 

The section of a beam where the bending moment is 
maximum is called the Dangerous Section^ and the prob- 



64 MECHANICS OF MATERIALS 

lem of finding tliis section is simply one of finding where 
tlie vertical shear passes through zero. 

Drawing the shear diagram, if the shear passes through 
zero under a concentrated load, the dangerous section 
is determined at once. When the shear becomes zero 
between two concentrated loads, the general equation for 
the vertical shear may be written for that part of the 
beam and equated to zero. 

The value of x which satisfies this equation determines 
the dangerous section. 

Having found the dangerous section, the bending 
moment may be calculated for that section, and when this 

value of M\^ substituted in the formula M= — , the value 

c 

of aS' will be the maximum unit stress in the beam. 

Art. 46. Relative Strengths of Simple and Cantilever 

Beams. 

Let the uniform load on either kind of a beam be ^y 
and a single concentrated load at the middle of a simple 
beam or at the end of a cantilever beam also be W\ then 
if a be some number whose value depends on the kind of 
a beam and the way in which it is loaded, the maximum 

bending moment for the beam may be expressed as 



a 



OJ 

This value substituted in the general formula M= — , 

c 

gives = - — , which may be written W= — — . The 

a c cl 

strength of a beam may be defined as the weight it will 

carry with a given unit stress. From the above equation 

for W it is evident that the weight a beam will carry with 

a given unit stress depends on the value of a, hence the 



BEAMS 65 

relative strengths of simple and cantilever beams loaded 
as above are directly proportional to «. If the expres- 
sions for the maximum bending moments in simple and 
cantilever beams loaded with W as above are written, an 
inspection of the results will show that for 

a cantilever beam loaded with W-dt the end a = 1, 

a cantilever beam loaded uniformly with W « = 2, 

a simple beam loaded with TFat the middle «= 4, 

a simple beam loaded uniformly with W «= 8. 

Art. 47. Overhanging Beams. 

Beams that overhang the supports are called Overhang- 
ing Beams. The fact that the reactions do not have their 
points of application at the ends of the beam does not 
prevent the application of the laws of mechanics to the de- 
termination of the magnitude of two reactions. Consider 
a beam that overhangs one or both supports and loaded in 
any way. Taking moments about a point in the line of 
action of one of the reactions as R^^ the moment of R^ — 
the moments of the loads on the left of R^ + the moment 
of the loads on the right of R^ = 0, and as the sum of all 
the forces is zero, R^ -{- R^ = the sum of all the loads. 

These two equations will suffice to fully determine the 
reactions R^ and R^ when the magnitude and position 
of the loads are known. 

The shear and moment diagrams can be drawn by the 
same principles that were applied to simple and cantilever 
beams. 

In general, the vertical shear will pass through zero 
at two or more points, giving more than one value of x 
for which the bending moment is a maximum. The value 
of the bending moment at each of these points must be 



66 MECHANICS OF MATERIALS 

calculated in order to find tlie greatest bending moment 
in the beam. 

The maximum moments may be either positive or 
negative, but when using the greatest value of M in the 

f ormula 31 = — , the substitution is to be made without 
c 

regard to sign. 

As the sign of the bending moment changes from posi- 
tive to negative, its value must pass through zero, and the 
position of tlie section of a beam for which the bending 
moment is zero is called an Inflection Point. The position 
of an inflection point may be approximately located by an 
inspection of the shear or moment diagrams and the general 
expression for the bending moment for that part of the 
beam written. 

Equating this expression to zero gives the position of 
the inflection point accurately. 

Art. 48. Beams of Uniform Strength. 

When the maximum unit stress in all sections of a beam 
is constant, the beam is said to be one of Unifoj^m Strength. 

The beams so far discussed have all had uniform sections, 

and the value of - was the same for all sections. 
c 

The bending moment iKf varies for all sections, and if S 

T ST 

is to be constant, - must vary with iHf, since 31 = For 

c c 

any beam loaded in any way the bending moment for a 
section at any distance from the left end may be expressed 
in terms of a variable distance x and this expression 

equated to — 
c 

Assigning different values to x^ the corresponding values 



BEAMS 67 

of - may be determined and the section of the beam at that 

' . I 

point chosen to satisfy the value of - as found above. 

c 

Beams of uniform strength may have any form of section, 
but they are usually made either rectangular in section or 

an approximation to such a section. 

. I hcl? 

For beams of rectangular section - equals —^ and 

ShdP' ... ^ . ^ 

M= ' in which either h or c? may be variable. Ex- 

6 ^ 

pressing Ji"in terms of the variable distance x^ the law of 
the variation of h or c?, as the case may be, determines the 
shape of the beam. 

At any section of the beam where 31 = 0, there is no 
moment to be resisted, and so far as bending is concerned 
the area of that section can be made zero. 

In addition to the unit stress due to bending, tliere is 
at all sections of the beam a shearing unit stress due to the 
vertical shear at that section. 

If S is the allowable unit stress in shear the area of the 

section where i^f = is given by J. = — -, where V is the 
vertical shear at that section. 

Art. 49. Moving Loads. 

In many cases the position of the loads is not fixed, and 
as the loads may occupy various positions, the value of 31 
for finding the greatest unit stress in a beam must be de- 
termined from the position of the loads which gives the 
greatest bending moment. 

When the loads on a beam may change their positions 
they are called moving or Live Loads to distinguish them 
from stationary or Dead Loads. 

Assume a beam, length Z, and Pj and P^ two unequal 
loads that pass over the beam. See Fig. 49. 



68 MECHANICS OF MATERIALS 

Let z be the distance of the greater load P^ from the left 
end of the beam, and jt? be the distance between the two 
Q loads. The dangerous sec- 

p— ^ * tion w^ill always occur under 

one of the loads, and assum- 
ing that it occurs under Pj 
M= B.^z. 



I 



^^^- *^- Let Q be the magnitude of 

the resultant of P^ and P^^ and x be the distance of its line 
of action from P^ ; then 

P^x = P^Qp - x^, 

^ ^ PiP ^ P3P. 

P^ + P^ Q 

Therefore the resultant of P^ and P^ acts at a distance 

P V 

z H ^ from the left end of the beam. 

Q 

Taking moments about a point in the line of action of 
the reaction R<^^ 

and R^z=Qz-^-?^^z = M. 



0, 







..,. ..- ^ 


I 


M will be 


a 


maximum when 








dM_ ^ 2Qz 
dz I 


l 


or 




1 P,^p 

ly si-i— . 

2 2$ 





Therefore when 3/ is a maximum the middle of the beam 
is halfway between the resultant of the loads and the 
dangerous section of the beam. 



BEAMS 69 

It is evident from the form of the equations that the 
same results would have been obtained had there been 
any number of forces. 

The results were obtained on the assumption that the 
dangerous section occurred under the first load, and if 
this assumption is true, the vertical shear must change 
sio-n as the section is taken to the rio^ht or left of the 
load Pj. 

When there are but two loads, the dangerous section al- 
ways occurs under the left hand load when the two loads are 
equal, and under the heavier load when they are not equal. 

When there are more than two loads, the position of 
the loads that gives the greatest bending moment does 
not always have the dangerous section under the maxi- 
mum load, but the general law holds true that 

When the middle of the beam is halfway hetiveen the re- 
sultant of the loads and the dangerous section a maximum 
bending moment occurs. 

Art. 50. Use of Formula. 

To find the position of a system of loads that causes 
the greatest bending moment, assume the loads to be 
so placed that when the dangerous section is assumed to 
occur under any load, the above criterion is satisfied and 
the vertical shear passes through zero, and calculate the 
iDcnding moment for that position. Tlie result will be 
the maximum bending" moment that can occur and have 
the dangerous section under the load as assumed. 

Assuming the dangerous section to occur under any 
other load, the maximum moment for that position may 
Le found. 

The position that gives the greatest value of M will be 



70 MECHANICS OF MATERIALS 

the one where the greatest bendmg moment occurs as 
the loads move over the beam. 

When a uniform live load moves over a beam, th& 
greatest bending moment occurs when the load extends 
over the entire length of the beam. When the load only 
partially covers the beam, the maximum moment occurs 
at the dangerous section and the above criterion holds true. 

The greatest vertical shear caused by any moving load 
is found at the supports when the resultant of the loads 
is nearest to that support. 

Art. 51. examination 

1. When is a bar called a beam? A simple beam? a 
cantilever beam ? a continuous beam ? 

2. What is meant by, " the reactions at the supports " ? 

3. Define the term Uniform Load; Concentrated Load. 

4. Name the laAvs of mechanics that are used to deter- 
mine the reactions for a simple beam. 

5. Define Vertical Shear and Bending Moment. 

6. As, " the sum of the forces," as well as, '' the sum of 
the moments of the forces," acting on each side of the sec- 
tion are equal, why is it necessary to use the expression, 
" acting to the left of the section," in giving the above 
definitions ? 

7. From the definitions of the bending moment and 
vertical shear, write the expression for each in terms of a 
variable distance from the left end of the beam. 

8. Why is it necessary to say, "moments being taken 
about a point in that section," in defining the bending 
moment at any section ? 

9. Define Resisting Shear; Resisting Moment. 



BEAMS 71 

10. Certain laws are deduced from experimental obser- 
vations made on beams under the action of bending forces. 
What are they ? 

11. What is the Neutral Surface of a beam ? the Neutral 
Axis of a section of a beam ? 

12. Show that the neutral axis passes through the center 
of gravity of the section. 

13. Prove that S=~: 

14. State clearly what each symbol in the equation 
S = ^—- means, and the units that should be used in sub- 
stituting for each. 

15. What conditions as to loads and material must any 

beam satisfy in order that aS'=^— - will be true for that 
beam ? 

16. What is a Modulus of Rupture ? 

17. Define the term Strength of a beam. 

18. Show that the strength of any beam depends on the 
value of «, where ct is a number depending on the kind of 
a beam and the nature of the loading. 

19. What is a Shear diagram ? a Moment diagram ? 

20. Define the Shear and Moment line. 

21. The shear line for a beam carrying only concentrated 
loads consists of horizontal and vertical straight lines. Why? 

22. When only uniform loads are considered, the shear 
line is a straight line from end to end. Why ? 

23. Show that for a beam with both concentrated and 
uniform loads, the moment line will be a series of curved 
lines. If there are no uniform loads, show that the 
moment line will consist of a succession of straight lines 
at different inclinations. 



Y2 MECHANICS OF MATERIALS 

24. If any part of the moment line is straight and 
parallel to the line representing the beam, what can you 
say of the bending moments for any sections taken in that 
part of the beam ? 

25. Define the expression, "the Dangerous Section of a 
beam." 

26. Give the relation that exists between the maximum 
bending moment for any beam, and the vertical sliear for 
that section. 

27. Show that the problem of finding the section of a 
beam where the bending moment is a maximum, is the 
same as that of finding the section where the vertical 
shear passes through zero. 

28. What are overhanging beams ? 

29. How do they differ from simple beams? 

30. Show why the reactions may be found in the same 
manner as for simple beams. 

31. AVhat is meant by the term Inflection Point ? 

32. If the vertical shear is zero at more than one sec- 
tion of a beam, how can you find the greatest bending 
moment for that beam ? 

33. When is a beam said to be one of Uniform Strength? 

34. Show that if the section of any beam is varied so 

that — varies with M^ the beam will be one of uniform 
c 

strength. 

35. When are the loads on a beam called Moving Loads? 

36. Give the criterion for the position of a system of 
moving loads that causes a maximum bending moment. 

37. If there is more than the one position of the loads 
that satisfies the criterion, how can you tell which position 
causes the largest bending moment? 



BEAMS 73 



PROBLEMS 



1. Find the reactions for a simple beam carrying a 
uniform load of w lb. /in. The length is I in. and the 
whole weight W lb. 

Solution. As the whole load is TF, the sum of the two reactions 
must equal W or p , r> _ tp- 

Taking moments about a point in the line of action of 7?.,, since the 
resultant of the uniform loads act at the center of the beam, 

Rd- — = 0, 
2 

H, = — = — and iin = — . 
^22 ^2 

2. Find the reactions for a simple beam, length Z, carry- 
ing a single concentrated load P at a distance p from the 
right support. 

Solution. The sum of the reactions equals the loads; hence 

B, + Bo = P- 
Taking moments about a point in the line of action of Ttg, 

R^l - Pp = 0, 



Substituting the value of iip 






^+R^ = P or R, = P(^1-B 

3. Find the reactions for a simple beam 10 ft. long 
carrying a uniform load of 500 lb. /ft. 

4. Find the reactions for a simple beam 30 ft. long 
carrying a load of 10 tons at a distance of 20 ft. from the 
left end of the beam. 

5. Find the reactions for a simple beam 30 ft. long 
carrying two equal loads of 5 tons each at 10 ft. from 
either end, and a uniform load of 500 lb. / ft. 



74 MECHANICS OF MATERIALS 

6. Find the vertical shears at the middle and at the 
ends of the beam in problem 3. 

Solution. From the definition of vertical shear, F at a section 
taken jnst to the right of R^ is the value of that reaction, or 2500 lb. 

At the middle of the beam the force acting upward is simply the 
left reaction, and the forces on the left of the section that act down 
are the unit loads on that part of the beam, or one half the total uni- 
form load. Therefore the vertical shear at the middle is 

2500 - 2.500 = 0. 

At the right end just to the left of the right reaction 

2500 - 5000 = - 2500 lb. 

or the vertical shear at iig is — 2500 lb. 

7. Find the vertical shears in the beam given in prob- 
lem 4, at sections taken just to the right and left of the 
load and at each end of the beam. 

8. Find the vertical shears in the beam in problem 5, 
at each end of the beam and at sections taken just to the 
right and left of each load. 

9. Find the bending moment at the middle of a simple 
beam, length I in., 

(a) for a load P at the middle. 

(6) for a uniform load of w lb. /in. 

P 
Solution for (a). The reactions are each—, hence by the defi- 
nition of the bending moment, for a section - from the left end of 

P I PI 
the beam ilf = — x - = — , as there are no other forces acting between 

2 2 4 
R^ and P. ^ <^ 4 

10. Find the bending moments at section just to the 
right and left of the load on the beam given in problem 2. 

11. Find the bending moment at the wall for a canti- 
lever beam, length I in., when the beam carries, 

(a) a uniform load of w lb. / in. 

(^) a single load P at the free end. 

(c) a single load P at jt? in. from the free end. 



BEAMS 



75 



Solution for (a). As there is no left reaction, the bending 
moment is the moment of the forces acting on the beam about a point 
in a section taken at the wall. The moments of all the uniform 
forces being equal to the moment of their resultant, let the resultant 
of the forces wl equal W; then 

M = -W-= -— . 

9 9 

12. Find the bending moment at each load and at the 
middle of the beam given in problem 5. 

13. A cast iron bar 1 by 1 in. in section and 36 in. long 
is broken as a beam. The modulus of rupture is 35,000 
Ib./sq. in. Required the maximum bending moment. 

14. Draw the shear dia- 
grams approximately to scale 
for simple beams 30 ft. in 
length loaded with 

(a) a uniform load of 100 
lb. /ft. 

(5) a concentrated load of 
3000 lb. at the middle. 

((?) a uniform load of 100 
lb. /ft. and a load of 3000 lb. at the middle. 

(c?) two equal concentrated loads at 5 and 10 ft. from 
the left end. 

Solution for (a). The vertical shear at any section distant x 



3000 

.ooooocxxxxxooo. 



t5oo 



t 



|l5< 





B 



Problem 14 a. 



n 



from the left end is 



V = 1500 - 100 X. 



This is the equation of a straight line, and as V is 1500 lb. at the 
left support and — 1500 lb. at the right support, if AB is 30 ft., AC 
= 1500 lb., and BD = — 1500 lb., then the line through the points 
C and D will be the shear line, and the diagram ABDC will be the 
shear diagram. 

15. Draw the shear diagram approximately to scale for 
a cantilever beam 10 ft. long loaded with 
(«) a uniform load of 500 lb. /ft. 
(6) a concentrated load of 500 lb. at the left end. 



76 



MECHANICS OF MATERIALS 




((?) a uniform load of 500 
lb. /ft. and a concentrated load 
of 5000 lb. at the free end. 

(d) two equal loads 2500 lb. 
each at 5 and 8 ft. from the free 
end of the beam. 



Problem 15 6. 



SoLUTiox FOR (b). The equation 
of the shear line is V= — 500, since there is no left reaction and the 
only load is 500 lb. at the end. 

Therefore, the shear line will be a straight line parallel to AB and 
at a distance — 500 from that line. A BCD is the shear diagram. 

16. Draw the moment diagrams approximately to scale 
for the beams as given in problems 14 and 15. 

Solution for the beam ix 14 a. The value of the bending 
moment at any section x from the left end is 



M 



Ji,x 

1 o 



^''-^ =1500 a: - 100 — 



where x is in feet. Giving any values to x, the corresponding value of 
M may be found. 



When X = 0, 

X = ft., 
a; = 10 ft., 
X = 15 ft.. 



M = 0. 

M = 6250 ft. lb. 
3/ = 10,000 ft. lb. 
M = 11,250 ft. lb. 



3000 

nrmnnnnnnno 



1500 



1500 



As the loads are symmetrical with the middle of the beam, the 
values of AI for x equal 20, 25, and 
30 ft. will be the same as for x 
equal 10, 5, and ft. Let AB = 
30 ft. and CD, EF, GH, etc., repre- 
sent on some scale the values of M 
corresponding to x equal 5, 10, 15 
... 30 ft. ; then a smooth curve 
passing through ADF, etc., will be 

the moment line, and the figure >* 5 io 15 &o £5 
A HE will be the moment diagram. Problem 10. 




BEAMS 77 

17. Write the expressions for the value of M and V 
for any section of the beam, and determine the maximum 
bending moments and vertical shears, 

(^d) for a simple beam loaded uniformly with W lb. 

(h') for a simple beam loaded at the middle with W lb. 

((?) for a cantilever beam loaded uniformly with W lb. 

(c?) for a cantilever beam loaded with W at the end. 

The length of all the beams being I ft. 

Solution for (a). The expressions for M and Fmay be obtained 
from the general formula by making " the concentrated loads to the 
left of the section " equal zero. 

Therefore, M = R,x '— = R,x , if W = wL 

2 2 

and 



v = 


R,- 


- tvx, 


3 

4 - 


U'l 

2* 




F = 


wl 


— wx. 



Hence 

2 

This expression is zero when x = -, and as M is a maximum when 

V is zero, substituting the value of x, which renders F = in the 
expression for M, gives 

II r _ wlx ICX^ _ WP Wl^ _ tvl^ _ Wl 

Evidently F is a maximum when x is zero. 

18. A simple beam is 20 ft. long and carries two con- 
centrated loads, one 100 lb. at 5 ft. and the other 500 lb. 
at 8 ft. from the left end, and a uniform load of 100 lb. /ft. 
extending over the beam for a distance of 12 ft. from the 
right end of the beam. Draw the shear and moment 
diagrams and calculate the maximum bending moment 
and vertical shear. 

19. A simple wooden beam 20 ft. long, 8 in. wide, 10 
in. deep, carries a uniform load of 80 lb. / ft. Required 
the maximum unit stress in the beam. 



78 MECHANICS OF MATERIALS 

Solution. For a uinformly loaded simple beam the maximum 

value of M is : 

Wl 80 X 20 X 20 X 12 .^f,^.. . ,, 

■ — = = 48,000 in. lb., 

8 8 ' 

d 

and ^ = 1- = ^= 6 ^ _6_ 

I hd^ hd-' 8 X 10'^ ~ 800' 
12 

e Mc 48,000 x 6 opa n / 

S= = — ^ = 360 lb. / sq. m. 

/ 800 ^ 

20. A simple wooden beam, rectangular in section, and 
20 ft. long, is to be designed to carry a load of 240 lb. at 
the middle with a maximum unit stress of 300 lb. / sq. in. 

{d may be assumed to be equal to 6 h.) 

Solution. The maximum bending moment is : 

,, PI 240 X 20 X 12 ,7 hd'^ d^ 
M = — = and - = — = — , 

c 6 36 



and 



M 

S 



4 








4 




I 

c 


.-. 


.1/ 


X 

S 


36 


= 6/3. 


240 


X 


20 


X 


12 


X 36 



Whence d^ = -^^ ^ -^ ^ ^- ^ -^^ ^ 1728, or ^ = 12, 6 = 2. 
4 X 300 ' 

21. A simple wooden beam is 1 foot square and 10 yd. 
long. What uniform load can it carry if the unit stress 
is not to exceed 300 lb. / sq. in. ? 

22. If the beam in problem 21 also carried a load of 
1000 lb. at the middle, what uniform load may also be 
carried if the unit stress is not to exceed 400 lb. / sq. in. ? 

23. Two planks, 12 in. wide and 2 in. thick, are placed 
one on top of the other and used as a simple beam. They 
support a uniform load of 1000 lb. What is the maximum 
unit stress in the material ? What would it be if the 
planks were placed side by side and carried the same load? 

24. A simple beam of wrought iron is 4 in. wide, 6 in. 
deep, 12 ft. long, and carries a uniform load of 32,000 lb. 
Is it safe ? 



BEAMS 79 

25. If the beam in problem 24 was a cantilever beam, 
what uniform load will it carry with a maximum unit 
stress not greater than the elastic limit ? 

26. A simple steel 10 in. I beam weighing 25 lb. / ft. 
is 18 ft. long and carries a uniform load, including its own 
weight of 15,000 lb. Required the maximum unit stress. 

Solution. The table gives the value of - for this beam as 24.4. 

The maximum value of M is ^ 

m ^ 15,000 X 18 X 12 ^j^^ s = — 

8 8 " / * 

rj c 15,000 X 18 X 12 X 1 i«AAAiK / o^ ,•„ 
Hence S = — = lo.oOO ib. / sq. m. 

8 X 24.4 ' ^ 

27. Show that when the weight TF of a simple beam is 
2 % of the load at the center, the error in the unit stress 
as found by neglecting the uniform load due to the weight 
of the beam is about 1 %. 

28. A common rule states that when the load at the 
center of a simple beam is greater than five times the 
weight of the beam the weight may be neglected when 
making the calculations for strength. What maximum 
error will this rule allow ? 

29. If W is the weight of a cantilever beam and P the 

P 
load at the end, find the ratio of -^ when the error in 

the unit stress caused by the neglect of TF is 5 %. 

30. Select a simple I beam of structural steel 24 ft. 
long to carry a load of 12,500 lb. at the middle with a 
factor of safety of 4. 

The maximum value of M is 

3^ ^ PZ ^ 12^ ^ 24 X 12. s = ^Jh^ = 15,000 Ib./sq. in. 
4 4 4 ^ 

M^I^ 12,500 X 24 X 12 ^ g^ 

S c 4x15,000 

The table gives the value of / for a 15 in. beam weighing 45 lb. /ft. 

as 60.8, and as P = 15 W, this beam will satisfy the conditions. 



80 mecha:n"ics of materials 

31. Select a standard steel I beam 10 ft. long to be used 
as a cantilever beam, to carry a load of 4 tons at the free 
end. Factor of safety 4. 

32. Select a simple steel I beam for a span of 20 ft. 
to carry two equal loads of 2 tons each at 5 ft. from either 
end and a uniform load of 300 lb. /ft., the unit stress 
not to be greater than 16,000 lb. /sq. in. 

33. If the term Coefficient of Strength is defined as the 
product of the total uniform load on a simple beam multiplied 

by the length of the beam in feet, show that - varies as this 
product. 

34. Show that the coefficient of strength for a simple 
beam carrying a concentrated load at the middle is twice 
that for the same load uniformly distributed. 

35. Find the relative values of the coefficient of strength 
for cantilever beams uniformly loaded, and loaded at the 
free end, compared with the same load uniformly distrib- 
uted over a simple beam. 

36. Select a standard steel channel 12 ft. long to be 
placed with the flanges vertical and used as a simple beam 
to carry a uniform load of 15,000 lb. Factor of safety 4. 

37. A beam 30 ft. long is supported at points 10 and 5 
ft. from the right and left ends. There is a uniform load 
of 500 lb. / ft. between the supports and concentrated 
loads of 450 lb. at either end of the beam. 

(a) Draw the shear and moment diagrams. 

(^) Find the greatest bending moments and vertical 
shears. 

(c) Find the inflection points. 

((?) Select a steel I beam to carry the loads with a 
maximum unit stress of 16,000 lb. /sq. in. 

38. A beam supported at two points 18. ft. apart over- 
hangs each support 6 ft. The overhanging ends carry a 



BEAMS ' 81 

uniform load of 300 lb. / ft. and there is a concentrated 
load of 12,000 lb. at the middle of the beam. 

(a) Draw the shear and moment diagrams. 

(5) Find the greatest bending moment and vertical 
shear. 

((?) Find the inflection points. 

(c?) Select a steel I beam to carry the loads with a 
maximum unit stress of 16,000 lb. / sq. in. 

39. Three men carry a stick of timber 12 x 12 in. x 12 
ft. long. One man is at one end and the other two are at 
such a point that each of the three men carries an equal 
load. Find that point. 

40. A cantilever beam of uniform strength rectangular 
in section is 12 ft. long and carries a load of 1200 lb. at 
the free end. The material is cast iron and the factor of 
safety is 10. 

Find the largest and smallest sections and make sketch 
showing the plan and elevation of the beam when, 
(a) the width is constant at 4 in. 

(5) the depth is constant at 12 in. 

41. A simple beam of uniform strength is rectangular 
in section and 12 ft. long and carries a uniform load of 
9600 lb. The material is cast iron and the factor of safety 
is 10. Find the smallest and largest sections and make a 
sketch showing the plan and elevation of the beam when, 

(a) the width is constant at 4 in. 

(6) the depth is constant at 12 in. 

42. If the beam in problem 38 was a rectangular steel 
beam of uniform strength and constant depth, find the 
proper size for the largest section when h = d for that sec- 
tion and the safe working unit stress is 16,000 lb. / sq. in. 
If the allowable unit stress in shear is 10,000 lb. /sq. in., 
find the area of the least section possible. Make a sketch 
of the plan of the beam. 



82 MECHANICS OF MATERIALS 

43. Two equal loads are 6 ft. apart. Find their position 
as tbey are moved over a simple beam 20 ft. long that 
gives the greatest bending moment in the beam. 

44. Three loads each 4 ft. apart are moved over a beam 
18 ft. long. From left to right the loads are 4000, 2000, 
and 2000 lb. Find the position of the loads that gives 
the greatest bending moment in the beam. 

45. As the loads (problem 44) pass over the beam from 
right to left, show that the maximum vertical shear for any 

position of the loads is given by F = ( 1 ]8000, where 

a is the distance of the 4000-lb. load from the left support, 
and b the distance of the resultant of the loads from the 
same load. 

46. As the loads pass over the beam (problem 44), 
make a diagram, using as ordinates the vertical shear to 
the left of the 4000-lb. load and as abscissa the distance 
of that load from the left end of the beam. 

47. If Q is the resultant of a system of loads moving 
over a simple beam, x the distance of Q from the left end, 
and a the distance of Q from the middle of the beam, 
show that the conditions for maximum moment require 

that J 

I — x=x— 2a when x >-, 

'A 

and , . n 1 ? 

I — x = x + Za when x <^'i 

and if this condition is satisfied, that the value of the 
maximum moment is given by 

yr Q ^ n2 f the moments of the loads on| 

-^^max — y V^ ± ^ «^ I ti^e left of the dangerous section. J 



CHAPTER IV 



TORSION 

Article 52. Derivation of Formula. 

In the previous chapters the forces were assumed to 
act in a plane passing through the axis and were either 
parallel or perpendicular to the axis. 

The forces that produce the stress known as torsion act 
in planes that are perpendicular to the axis of the bar, 
and while the lines of action of the forces are perpen- 
dicular to it, they do not pass through the axis. 

The effect of such forces must be to twist the bar. 
Assume a cylindrical bar, one end of which is firmly 
fixed in the wall, to be 
acted on by a couple ly- 
ing in a plane perpendic- 
ular to the axis of the bar 
at a distance I from the 
wall, and whose moment 
about that axis is Pp. 
A fiber of the bar that 
before the application of the force occupied the position 
of the line ad (Fig. 52 a), after the force has been applied 
will occupy the position of the helix ab^ and a point d on 
the surface of the bar will have moved to the position h. 

It is evident that the angle had^ the angle of the helix, 
is independent of the length of the bar and depends only 

83 




Fig. 52 a. 



84 



MECHANICS OF MATERIALS 



on the twisting forces and the material of the bar. Since 
any plane section perpendicular to the axis of the bar 
between the wall and the couple would contain an arc 
similar to hd^ and the length of that arc is proportional 
to the distance from the wall, the angle hod is propor- 
tional to the length of the bar and the twisting forces. 

This angle is called the angle of twist, and will be 
denoted by 0. 

By analogy with tension, the distance a point on the 
end of the bar moves under the action of the twisting 
forces being similar to the distance a point on the end 
of a tension bar moves under the action of the tensile 
forces, the arc hd may be taken as a measure of the 
deformation of the surface fibers of the bar due to the 
twisting forces. 

As the arc hd was proportional to the length, — can 
be taken as representing the unit deformation. 

Experiment has proven that when a bar is circular in 
section and no stress is greater than the elastic limit, the 

line od^ moved to any 
new position during the 
twisting of the bar, re- 
mains a straight line. 
This being true, the 
length of any arc d^-^ 
(Fig. 52 5), with a center 
at and a radius y^ is- 
proportional to its radius, 
and as the arc is propor- 
tional to the deformation 
at the radius y, the force 
Fig. 52 6. producing this deforma- 




TORSIOX 85 

tion is proportional to tlie same distance. This reasoning 
is true for any perpendicular section of the bar between 
the plane of the couple and the wall. 

Assume the bar to be cut by a plane perpendicular to 
the axis between the wall and the twisting couple, and 
introduce forces in that section to render the free end in 
equilibrium. These forces must all act in a plane per- 
pendicular to the axis, since the external couple has no 
component perpendicular to such a plane, and their re- 
sultant must be equivalent to a couple whose moment is 
equal to that of the twisting moment. 

Considering the two faces of the bar in any section, 
it is evident that the face on the right end tends to slide 
against the face on the left end as the former tends to 
rotate about the axis of the bar, thus producing a shearing 
stress throughout the section. Since the force, and con- 
sequently the unit shearing stress, is proportional to the 
distance from the center of the bar, if we let S^ be the 
unit shearing stress in the surface fibers, e be the distance 
of those fibers from the axis of the bar, and ?/ be the dis- 

tance of any fiber from the same axis, then — is the unit 

c 

stress at a unit's distance from the axis and — ^«/ is the 

c 
unit stress at any distance y. 

Considering the stress uniformly distributed over any 

small area dA at the distance ?/, — i/dA is the force acting 

c 

on that elementary fiber, ^ml ^-^ y^dA is the moment of 

c 

this force about the axis of the bar. But the sum of the 
moments of the forces acting in the section is equal to the 

twisting moment Pj9, hence — j i/^dA = Pp. f if'dA is 



86 MECHANICS OF MATERIALS 

the expression for the polar moment of inertia about an 
axis through the center of gravity of the section, and 
writing J for ^ ^j 

I 'if'dA gives Pp = — ^, (e) 

which shows the relation between the maximum unit stress 
in shear and the twisting forces. 

Art. 53. Modulus of Section. 

ST 

Comparing formula (e) with M= — , it will be noticed 

c 

that they are of the same general form, iltfand Pp repre- 
sent the effect of the external forces, S in each case is the 
maximum unit stress in the section, and by analogy with 

-, - may be called a modulus of the section. Formula 

c c 

(e) is only true for bars whose sections are circular, and 
where the material and loading of the bar satisfies the 

conditions stated for M= — 

c 

Art. 54. Square Sections. 

For rectangular sections the formula is only ajoproxi- 
mately true, as a radial line drawn from the corner does 
not remain straight during the twisting of the bar, as was 
the case with the circular section. 

The investigations of St. Venant cover the rectangular 
section, and his results give for a rectangular shaft sub- 
jected to torsion, g 

^i? = ^.j-g (nearly), 

in which d is the side of the square. From the form of 

J d^ 

the expression the effective value of - must be — — instead 

c 4.8 



TORSION 87 

of the calculated value , hence the value of Pp as 

found from the above formula is less than that obtained 
from equation (e), when equal values of S^ and d are 
used. 

Square sections are apt to be weaker than St. Venant's 
formula would indicate, since the maximum stress is 
carried on the edge of the square and any slight defect 
reduces the effective diameter of the bar. 

For this reason square sections are rarely used to resist 
torsion alone. 

Art. 55. Illustrations. 

The resistance at the wall may be assumed as another 
couple, whose moment is equal and opposite to that of 
the twisting moment, without altering the conditions as 
assumed when formula (e) was developed. In the case 
of a line shaft, where the belt from the engine produces 
a twisting moment at one end of the shaft, the resistance 
of the belt on the pulley at the other end is equivalent 
to an equal moment. If there are several pulleys on the 
same shaft, each producing a moment by its resistance 
to turning, it is evident that the resisting moment of the 
shaft will not be constant throughout the length, but will 
vary with the resistance that it has to overcome. To 
illustrate, assume a shaft with three pulleys, one at each 
end and one at the middle of the shaft. The driving 
moment is applied at the left end and the resisting 
moment of the other two pulleys, PiPi and P^f^', must be 
equal to the driving moment Pjp, The moment to be 
resisted by the shaft between the driving pulley and the 
one at the middle is P]p = P^Vx + ^iVv After passing 



88 MECHANICS OF MATERIALS 

the middle pulley the resistance to be transmitted b}^ the 
shaft is only the twisting force of the third pulley, and 
consequently the resisting moment between the second 
and third pulleys is equal to P^p^- 

Art. 56. Twist of Shafts. 

In Fig. 525, dh was taken as a measure of the defor- 
mation of the surface fiber, and — the unit deformation 

of that fiber. From the figure dh = 6c and by the deh- 
nition of the modulus of elasticity, if S^ is the unit stress 
in the surface fibers, the shearing modulus of elasticity 

must be ^ ^i t> i 

j^_S,_SJ Ppl 

I 
The latter expression is obtained by substituting for aS'^ 

its value from S, = ^-• 

Equation (/) gives the relation between the modulus 
of elasticity for shear, the twisting moment or the unit 
shearing stress, and the angle of twist. 

When the data given in any problem is sufficient to 

determine two of the three quantities, the twisting moment, 

the unit shearing stress, or the section modulus, the 

V 'T 
formula Pp = ^' will completely determine the other 
c 

one. Or when the given data will determine three of 
the following quantities, the twisting moment or the unit 
shearing stress, the dimensions of the bar, the angle of 
twist, and the shearing modulus of elasticity, formula (/) 
€an be used to solve any problem involving the twisting 
moment or the unit stress and the angle of twist. 



TORSION 89 

Since all of the tabulated values of the constants of 
materials are given in pounds and inches, all dimensions 
of weight and linear or square measure must be reduced 
to pounds and inches before making the substitutions in 
the formulas. As the value of 6 used in the development 
of the formula was in circular measure, 6 must be ex- 
pressed in radians. 

Art. 57. Relative Strengths and Stiffness of Shafts. 

The strength of a shaft may be defined as the twisting 
moment it will carry with a given unit stress. Formula 
(e) shows that the strengths of shafts vary directly as 

the value of ^-i^ for each shaft, and when the shafts are 

^ . J 

of the same material, as — 

c 

Defining the stiffness of a shaft as the angle of twist 

for a given value of Pjt>, since = -^ or — ^, it is evident 

that the stiffness of two shafts of the same material varies 

directly as — or -, depending on whether the twisting 
J c 

force is given directly or in terms of the stress. 

Art. 58. Horse Power of Shafts. 

A horse power being defined as 33,000 ft. -lb. per 
minute, if H is the horse power to be delivered by a shaft 
making N revolutions per minute, the value of Pp in 
terms of the horse power may be found from the equality 
of tlie work done by the twisting force per minute and 
the work represented by Ahorse powers. Assuming that 
-P is a force acting at a radius p, the work done by that 
force in one revolution must be ^irPp in. -lb. and as 
a horse power is 33,000 ft.db. per minute, or 396,000 



90 



MECHANICS OF MATERIALS 



in. -lb. per minute, 2 7rP/j>iV = 396,000 H^ wliicli re- 



duces to H=^r, 



PpN 



'•' {U\Ci ('"^PPi'oximately). Substituting for 
Pp its value in terms of S^ from formula (e), gives 
H= ^^ — These two expressions may be used to de- 
termine the horse power that a given shaft will transmit 
when the number of revolutions per minute and the twist- 
ing moment or the maximum unit stress are known. 
The values of either Pp or Sg may be found from (/) 
and the angle of twist for a given horse power determined. 

Art. 59. Shaft Couplings. 

When two lengths of shafting are to be joined together, 
the connection is often made as in Fig". 59. The moment 



E:;: 



E 



E 



? 



:::;} 



J 




Fig. 59. 

of the shearing stresses in the bolts must be equal to the 
twisting moment of the shaft, and the relation between 
the twisting moment of the shaft and the resisting 

moment of the bolts may be stated as Pp = SJ — —-, 

where J' is the polar moment of inertia of the section of 
a bolt about the axis of the shaft, n the number of bolts, 
SJ tlie maximum unit stress in the bolts, and c' the dis- 
tance of the most distant fiber of the bolt from the 
axis of the shaft. 



TORSION 91 

If the bolts and shaft are of the same material, then 

- = '—J-, where e/ancl c refer to the shaft and J' and c' to 
c c 

the bolts. The polar moment of inertia of the bolt about 
the axis of the shaft is equal to J' = Jy-\- Ah^, where J^ 
is the polar moment of the bolt section about an axis 
through its center of gravity, and h is the distance of the 
axis of the bolt to the axis of the shaft. Expressing 

'_ and - — - in terms of the diameters of the shaft and 
e c' 

bolts, gives a method for finding the proper diameter of 
the bolts to be used. The equation thus formed gives a 
very awkward expression for the value of the diameter 
of the bolts, and it is common practice to assume that the 
shearing stress is uniformly distributed over the area of 
the bolts. This is equivalent to assuming that the result- 
ant of the shearing stresses acts at the center of the bolts. 
Let SJ^ be the unifoi-m unit shearing stress in the bolts, 
h the radius of the bolt circle, or the distance between the 
center of the shaft and the center of the bolts, and d the 
diameter of the bolts; then the resisting moment of 

Trd^ 
the bolts must be n-^-j-SJ' 7i, and equating this to the 

twisting moment of the shaft gives 

which is in convenient form for use in the determination 
of the bolt diameters. 

As the resisting moment of the bolts must be equal to 
the resisting moment of the shaft, if S^ is the maximum 
unit stress in the shaft and D the diameter, then 



92 MECHANICS OF MATERIALS 

gives the relation between the unit shearing stresses in the 
shafts and bolts. Either the number of the bolts, their 
diameter, or the radius of the bolt circle can be taken as 
unknown. 

In applying the approximate solution to the solution of 

a problem, the total bolt area, n—r-^ needed, may be found 

and the number of bolts chosen that will give the required 
area. The assumption made in the approximate solution 

is also equivalent to assuming that J' = — - — and c' = h. 

J' 

Comparing the exact and approximate values of — -, we tind 

c' 

that the approximate value is nearly 10 % larger than the 
accurate one, when 7i = 4 c?. The error will increase as 
the ratio of A to c? increases, and of course decrease as that 
ratio decreases. The error, while not of much importance 
in the majority of cases, should always be considered wlien 
the decision is made regarding the diameters, the total 
bolt area as found by the approximate solution being less 
than the true area. 

Art. 60. Modulus of Rupture in Torsion. 

When the twisting moment Pp is great enough to rup- 
ture the bar, the value of S^ as found by equating this 

value of Pp to — has been called the modulus of rupture 

in torsion. Tlie value of such a constant of material is 
doubtful, as the formula is not true when the unit stress 
is greater than the elastic limit. It would seem to be 
better practice to use the maximum unit stress in shear, 
as determined by the use of P = ^aS' as the maximum unit 
stress in torsion, and base any factor of safety on that 
constant. 



TORSION 



93 



Art. 61. Helical Springs. 

When a wire is wound around a cylinder so that the 
axis of the wire forms a helix, the resulting form of the 
wire is called a helical spring. Take such a spring made 
from wire whose diameter 
is d and wound on a cyl- 
inder so that the diame- 
ter of the helix is i>, 
compressed or extended 
by tlie force P whose line 
of action passes through 
the centers of all the coils, 
and consider one coil. If 
the spring is closely 
wound and D is large compared with d^ the plane of the 
coil will be nearly perpendicular to the line of action of 

the force; hence the force P acting with a lever arm — 
tends to twist the wire. 





Fig. 61. 



The twisting" moment is 



PD 



and the section modulus 



^3 " pjy 

of the wire tt— ;; hence, taking aS'^ as the unit stress = 

lb 2 

IT — - Sg^ or P = — — aS'^ gives the relation between the ten- 
sile or compressive force P and the dimensions of the 
spring. 

The length of one coil of the spring is approximately 
ttD, and if the wire is twisted through a small angle 0, by 

PD 

a moment that increases uniformly from to —^^ the 

PD '^ 

work done on the wire of one coil is 6 , where is in 

circular measure. If we let A be the total deflection of 
the spring, that is the amount of shortening or lengthening, 



94 MECHANICS OF MATERIALS 

and B the deflection for one coil, as tlie force acting varied 

BP 

uniformly from to P, the work done on one coil is 

Since these two expressions represent the same quantity 

PD SP ^8 

of work, they must be equal, or — ■ — = — -- or 6 =^. 

As 6 is the angle of twist, formula (/) gives as its value 

^^ Ppl ^ 16 P^P 
FJ Fd"^ 

Equating the two values of ^, 

lb = — or o = • , 

Fd^ B Fd^ ' 

which is the deflection of one coil in terms of the load P. 
Taking Q^SJ^^h^S^rr^ 

Fc D Fd ' 

we have 3 = -^ as the relation between the unit stress 

Fd 

and the deflection for one coil. 

Since the strength of any bar under the action of twist- 
ing forces is independent of the length of the bar, the 
formula for the strength of a spring is also independent of 
the number of coils. If n is the number of coils, the total 
deflection of any spring must be n times the deflection for 

one coil, or ^Pmn S^irPhi 

A = or -^ ' 

Fd^ Fd 

EXAMINATION 

1. What is a torsional force ? How does it differ from 
a force that produces a shearing stress that is uniform over 
the section of a bar ? 

2. Can a torsional force produce any tensile or com- 
pressive stresses ? 



TORSIOX 95 

3. " It is evident that the angle had is independent of 
the length of the bar." Prove it. 

4. Show that the angle of twist depends on the length 
of the bar. 

5. If a straight radial line drawn on the end of a bar 
remains straight when the bar is being twisted, show that 
this fact proves that the unit stress is proportional to the 
distance from the axis. 

6. Show that S,= ^- 

J 

7. State under what conditions of load and material 

O J 

the formula Pp = -^ is true. 

c 

8. Define the terms '' strength of a shaft"; "stiffness 
of a shaft." 

9. Does the strength of a shaft depend on its length ? 
Is the same true of the stiffness ? 

10. Two shafts of the same diameter and lens^th are of 
different materials. What is their relative strength ? 
What is their relative stiffness ? 

11. Show how to find the expression for the modulus of 
elasticity in shear. 

12. Why was it necessary to reduce the value of a horse 
power, 33,000 ft. -lb., to inch-pounds in order to obtain the 
expression ^^JTV^ ^ 

~ 63,000 c" 

PROBLEMS 

1. One end of a circular bar 2 in. in diameter and 10 ft. 

long is fixed in a wall, and at the other end there is a 

couple Avhose moment is 300 ft. -lb. Required the unit 

stress in the bar ? 

Solution. The value of J —^^-— and c = — ; hence - — '^^—^. 
Pp = 300 ft.-lb. = 3G00 iu.-lb. '^^ ^ c 1(J 



96 MECHANICS OF MATERIALS 

S J 
Substituting these values in Pp = - — , and solving for Ss, gives 

c 

c, 3600 X 16 nor.A ir. / 

5s = : — = 2290 Ib./sq. in. 

TT X 8 

2. A circular bar 7 in. in diameter, 10 ft. long, is acted 
on by a force of 10 tons perpendicular to, and at a distance 
of 3.14 ft. from the axis. Required the unit stress induced. 

3. Find the diameter of a circular steel bar to carry a 
twisting moment of 20 ft. -tons. 

4. Find load that can be applied at the end of an arm 6 
ft. long so that the maximum unit stress induced in a cir- 
cular bar 2 in. in diameter will not exceed 12,000 Ib./sq. in. 

5. A circular steel bar 2 in. in diameter is twisted by a 
force at the end of an arm 6 ft. long. Required the force if 
the unit stress due to torsion is equal to the elastic limit. 

6. If the bar in problem 1 was soft steel, find the angle 
of twist. 

Solution. Taking the values as found for 1 and substituting in 

= —> Z = 10 X 12 in. and F = 12.000,000, 
Fc 

^ ^ 2290 X 10 X 12 ^ 22,900 ^ ,,29 radian, 

12,000,000 X 1 1,000,000 

or approximately 1° 18'. 

7. A soft steel bar is 6 in. in diameter and 20 ft. long. 
What force acting tangent to the surface will twist the 
bar through an angle of 1° ? 

8. A steel shaft 2 in. in diameter and 50 ft. long is 
transmitting a torsional moment that causes a unit stress 
equal to the elastic limit. Required the angle one end 
is twisted through relative to the other. 

9. Find the diameter of a steel shaft 10 ft. long to 
carry a twisting moment of 81,700 ft. -lb., if the unit 
stress is not to be greater than 10,000 Ib./sq. in., and 
the angle of twist less than 1.15°. 



TORSION 97 

10. Find the diameter of a steel shaft making 100 
revs. /min. and transmitting 200 H.P., the unit stress 
being 6300 lb. /sq. in. 

S.JN 6300 X TT X ^3 X 1 00 onn 

Solution. H = ^./^..^ = ^o r^ncx — T^ ~ '^^^' 

Dd,000 c 00,000 X lb 

,o 200 X 63,000 X 16 . ^^ . 
d^ = _ : . = 4.0/ in. 

6300 X TT X 100 
The shaft chosen would be either 4^ or 5 in. in diameter. 

11. The Allis Chalmers Co. base their tables for the 
strength of mild steel shafting on the formula 11= cd^ N^ 
where d is the diameter of the shaft in inches, and c a 
number which has the following values: 

For heavy or main shafts e = .008 

For shafts carrying gears c = .010 

For light shafts carrying pulleys c= .013 

Find the unit working stress allowable in each case. 

12. 6000 H.P. is to be transmitted through a shaft. If 
the shaft is a hollow cylinder 36 in. outside diameter, find 
the inside diameter. Take the unit stress as 12,000 
lb. / sq. in. and the revolutions per minute as 90. 

13. A coupling is to be used to connect two lengths of 
shafting 4 in. in diameter. The maximum allowable unit 
stress in the shaft is 10,000 lb. /sq. in., the diameter of 
the bolt circle is 6 in., and the allowable unit stress in the 
bolts is 8000 lb. /sq. in. Find the diameter and number 
of bolts necessary, assuming that the shear is uniformly 
distributed over the section of the bolts. 

14. A hollow shaft has the outside diameter twice the 
inside diameter. Compare its strength with that of a 
solid shaft of the same material and section area. 

15. If the elastic limit of the material in one shaft is 
60,000 lb. /sq. in. and costs 10^ per pound, what can 



9S MECHANICS OF MATERIALS 

3'ou afford to irdj for a shaft to do tlie same work, if the 
ehistic Ihuit of the material is 30,000 lb. /sq. in. ? 

16. A helical spring is made of wire w^hose diameter is 
1 in., the mean diameter of the coil 4 in., and has 30 coils. 
If the value of F is 12,000,000 and the working unit 
stress 60,000 lb. / sq. in., required the load it will carry 
and the deflection under that load. 

17. A helical spring has to carry a load of 5890 lb. 
and has a deflection under that load approximately 10 in. 
The unit stress, diameter of the wire, number of coils and 
the modulus of elasticity in shear are the same as in prob- 
lem 16: find the diameter of the coils. 

18. D. K. Clark gives as the deflection for one coil of a 
helical spring B = o,^ .4 , where d is the diameter of the 

wire in sixteenths of an inch. Assuming the value of F 
as 12,000,000, what unit stress does this formula allow? 



CHAPTER V 



THE ELASTIC CURVE 

Article 62. Definition. 

When the beam is bent, the neutral surface assumes 
a curved form; and the projection of this surface on a 
vertical plane parallel to the axis of the beam is called 
the Elastic Curve. If the equation of this curve for any 
beam is expressed as y=f{x)^ where y is the deflection 
of the beam at any point distant x from the left end of 
the beam, the deflection of the beam at any point can be 
easily found. 

Art. 63. Equation of the Elastic Curve. 

To derive this 
equation of the 
elastic curve let 
Fig. 63 represent 
a portion of a bent 
beam. 

Let y^ measured 
along the axis of 
the beam be tak- 
en as representing 
c??, where I is the 
length of the beam 
and the curve dl a 

differential part of Fig. 63. 

99 




100 MECHANICS OF MATERIALS 

the elastic curve, then if o is the center of curvature, oi 

and oj are radii. Before any bending took place these radii 

were parallel to each other, and eh may be assumed to have 

been in the position kd parallel to af. Let the greatest 

distance of the neutral axis to the surface fiber be c and 

assume that jh = c. The deformation of the outer fiber 

kb 
is kb and the unit deformation is — , since kb is the chano-e 

dl "^ 

in the length dl; ^ is the modulus of elasticity, and S 

the unit stress in that fiber, hence kb = . From the 

U 
similar triangles jkb and oij 

ji oj 
But oj = r = the radius of curvature, ji = dl, jb = c, and 
kb = -TT ; hence, substituting these values in (1), we have 
Sdl _ c S _c ,,o_ -^^^ 

ri 77 — ~i OV ~ — -; but O — —r- 1 

Edl r E r 1 

. M E EI .ox 

hence — = — , or r = , (1) 

It M ^ ^ 

is the equation of the elastic curve of a beam in terms 
of the modulus of elasticity, the bending moment at any 
point of the beam, the moment of inertia, and the radius 
of curvature of the elastic curve at that point. 

The value of r expressed in rectangular coordinates is 



r = g (3) 

dx^ 

Since the degree of curvature of any beam in an en- 
gineering structure is very small, the value of the tangent 



THE ELASTIC CURVE 101 

of the angle which the tangent at any point on the curve 
makes with the axis of X is a very small quantity and 

-^ ) may be neglected in comparison with unity. Equa- 
tion (3) then reduces to 

If dx is assumed to be equal to dl^ an assumption which 
is approximately true when the degree of curvature is 
small, and the value of r as just found is inserted in 
equation (2), we have 

dx^ 

which is the differential equation of the elastic curve of 
any beam. M is the bending moment for any part of 
the beam for w^hich the equation represents the curve, 
therefore must be expressed in terms of x any distance 
from the left end of the beam, and y is the ordinate to 
the curve or the deflection of the beam. 

Since the condition that ^ = — was introduced in the 

e 

•derivation, the unit stress must be within the elastic limit, 

3Ic 

and as the formula aS' = — — - was also used, all the condi- 
tions of the materials that are necessary to the correct 
use of that formula obtain with the one just developed. 
The assumptions that dl = dx^ and that the maximum 

value of — ^ was so small that [—-] might be neglected as 
dx \dxj 

compared to unity, introduce errors that, while they are 

small, increase as the degree of curvature increases. The 



102 MECHANICS OF MATERIALS 

use of the formula under ordinary engineering conditions 
gives results that agree well with those obtained in exper- 
imental work, the discussion of the limitations to its use 
being given to show that the formula is not applicable to 
all cases. 

Art. 64. Deflection of Beams. 

Let 31 = f{x) be the expression for the bending moment 
for a portion of any beam loaded in any way; then 

^jfl =/(:.) (1) 

is the differential equation of the elastic curve for that 

part of the beam for which /(rr) represents the bending 

moment. 

Integrating (1), 

^J^=/'(x)+C, (2) 

ax 

where C is a constant of integration. The value of 

may generally be found by noting that --^ is the tangent 

of the angle which the tangent at x makes with the axis 
of JT, and from the conditions of the problem, finding a 

value of X for which -^ is either zero or known. 

dx 

Integrating again, 

UIi/=f'(x') + Cx+ C,. (3) 

The value of 6\, the constant of integration, can in 
most cases be determined by finding a value for x for 
which ^ is either zero or known. 

When O and 0^ are determined, equation (3) may be 
used to find the deflection at any point of the beam for 
which the bending moment is equal to /(^). 



THE ELASTIC CURVE lOa 

As the right hand member of equation (2) is the differ- 
ential coefficient of equation (3), the value of x which ren- 
ders f {x) 4- (7 = makes the value of ?/ a maximum, or 
the maximum deflection for that part of the beam may be 
found. As an aid in the determination of the constants 
of integration, the student is reminded that when the 
moment diagram is symmetrical about a vertical line at the 

middle of the beam, — ^ is zero for a value of x = -^ and 

dx 2 

for the special case of cantilever beams — ^ is zero at the 

dx 

wall, as the restrainment keeps that part of the beam 

fixed and horizontal. 

Since there is no deflection at the supports, y will always 
be zero at the points where the beam is supported. When 
there are concentrated loads on the beam, the expression 
for ifer will take a different form for each part of the beam 
between the loads or between the loads and reactions. 

If there are n concentrated loads on the beam, there 

will be 9i + 1 forms that the ex23ression for the bending 

moment may take, making n -\-\ equations of the elastic 

curve, tlie double integration of each bringing into the 

problem 2 (ii + 1) constants of integration. 

For simple beams the value of x that will make -^ = 

dx 

is not known unless the loads are symmetrical with the 
middle of the beam, but y is always equal to zero at the 
supports where x is either or I. 

Any two of the n •\-\ equations representing consecu- 
tive portions of the beam will have the expressions for -^ 

dx 

and y equal for a value of x at the load where the equa- 
tions meet. As there are but 2 ri + 2 constants of inte- 



104 MECHANICS OF MATERIALS 

gration, and two known conditions are always to be had 
from the fact that 3/ = when x= or x = I, the 2 n 
equations resulting from the equating of the values of 

— ^ and ^ under each load will suffice to determine the 
dx 

other 2 n constants of integration. 

In using equation (3) for the determination of the 
maximum deflection of a beam, the student is again 
reminded that the equation only holds true for that 
portion of the beam represented by /(a;), and where 
there are several concentrated loads, each portion may 
have to be investigated, in order to find the greatest 
value of t/ for the beam. In general, an inspection of 
the distribution of the loads on the beam will show the 
portion of the beam where the greatest deflection is likely 
to occur. 

If the bending moments for simple and cantilever beams 
loaded uniformly with TFJ simple beams loaded at the 
middle with IT, and cantilever beams loaded with W at the 
end, are expressed in terms of 2:, these moments may be 

substituted in UI — ^■ = M^ forming four differential equa- 
dx^ 

tions. Integrating each equation twice gives the value of 

y^ the deflection of the beam for any value of x. The 

maximum value of the deflection y may be expressed as 

1 Wl^ 
f z=z —^^ where yQ is a constant depending on the kind 

of a beam and the nature of the loads. Transposing, 

TpTf 

W= B—-^. The load a similar beam will carry was 

73 "^ 

SI 

given as ir= « — , and if these two expressions for W 
Ic 



aSP 



are equated we find that f = — — -, giving the relation 

l3 Ec 



THE ELASTIC CURVE 105 

between the maximum deflection, the dimensions of the 
beam, the unit stress, and the modulus of elasticity. 

Attention is called to the way / varies in the two 
expressions for the maximum deflection. 

When the load W is considered, / varies directly as P 
and inversely as Z, and when the unit stress is given, / 
varies directly as P and inversely as c. For rectangular 
sections, this latter variation makes the maximum deflec- 
tion independent of the breadth of the beam. 

Aet. 65. Restrained or Fixed Beams. 

A beam is said to be restrained or fixed when one or 
both ends are so firmly imbedded in the wall that the tan- 
gent to the elastic curve at the fixed ends always remains 
horizontal during the flexure of the beam. A cantilever 
beam under this definition is a fixed beam when it projects 
from the wall ; but since there was no reaction at the free 
€nd, the bending moments and shears can be determined 
without reference to the fixed end. When the free end of 
a cantilever beam has a support placed under it, the con- 
ditions are changed. The magnitude of the reactions 
can not be determined from the conditions for mechanical 
equilibrium that were used for simple beams. The value 
of Jf for any section of the beam will contain a term that 
includes the unknown reaction, and this value of M sub- 
stituted in the general equation of the elastic curve will 
give the differential equation of the curve for this beam. 
When this equation is integrated twice, the value of the 
unknown reaction may be found by applying the known 
conditions that the resulting equations must satisfy. 

Let Fig. 65 represent a beam fixed at one end and 
supported at the other, loaded in any way. If Mi is the 



106 



MECHANICS OF MATERIALS 



reaction at the left end of the beam, the bending moment 
for a section distant x from that end is 



wx 



2 rthe sum of the moments of the loads 1 

i!f"= Hy"^ \ to the left of the section with refer- U 

^ [ence to a point in that section J 

and putting this value of M\w. the general equation of the 
elastic curve gives 

rrr d y -wy ivx j sum of the moments ) 

d7? ^ 2 I of the loads, etc. \ ' 

which is the general equation of the elastic curve for 
a beam fixed at one end and supported at the other. 

If there are no concen- 
trated loads, the last 
term of the right hand 
member will be zero^ 
and if there is no uni- 
wx^ 



chrrhc'frhriMm'/. 



ir 



X 




form load, 



will be 




zero. In a previous 
article it was shown 
that in a general case 
there were 2 (n — 1) 
conditions for the deter- 

FiG. 65. • I.- £ 1 

mmation oi an equal 
number of integration constants, and in this case there is 

the additional condition that -^ = when x = l^ which 

ax 

may be used to determine jR^, since the tangent to the 
curve is horizontal at the wall. Stated briefly, the con- 
ditions that ?/ = at the fixed and supported ends, — ^ = 

ax 

at the fixed end, and the 2 n equations resulting from the 



THE ELASTIC CURVE 



107 



equating of the values of ^ and y under each load, will 

be sufficient to completely determine the values of the 
2 (9^ — 1) constants of integration, and that of the 
unknown reaction at the left end of the beam. 

Art. 66. Beams fixed at Both Ends. 

Let Fig. %'<S represent a beam fixed at both ends and 
loaded in any way. The forces which act between the 
beam and the walls supporting the beam, keeping the tan- 
gent to the elastic curve at the wall horizontal, are un- 
known. The un- 
known systems of '^'^ 
forces acting in 
each wall may each "A/// A^f 
be replaced by a y/a^ ^ 
single vertical 
force acting up- 
ward at the face of 
each wall, and a 
couple whose mo- 
ment is sufficient 
to keep the tangent 
at the wall horizon- 
tal. The beam, 
under the action of 
these forces and the 
loads, may then be considered as a body in equilibrium. 
From the mechanics of equilibrium of parallel forces, it is 
evident that the sum of the two vertical forces acting up- 
ward must be equal to the sum of the loads on the beam, 
and if the moments of the couples are determined, the 
values of the two vertical forces mav be found. 




Fig. 6G. 



108 MECHANICS OF MATERIALS 

Let jKj, (Fig. 66") be the force at the left end and M2 the 
force at the right end, and the moments of the couples at 
the left and right ends of the beam be iff^ and M^ respec- 
tively. Since the moment of a couple about any point is 
constant, the moment of the couple on the left end about 
a point in a section distant x from the left end is 31^^, and 
the value of the bending moment for that section is 

71^ Tlf" _i_ 7? "^"^^ ( ^^^® ^^^^ ^^ ^^^ moments ) 

This value of il[f substituted in UI^^ = Mis the differ- 

ential equation of the elastic curve for a beam fixed at 
both ends. If there are no concentrated loads, the last 
term of the expression for M will be zero, and when 
there is no uniform load, the term containing w will 
disappear. If there are n concentrated loads, there will 
be ^ 4- 1 values for M^ and each expression will contain 
the unknown moment M^. The double integration of the 
n-\-l equations will bring into the problem 2(?i + 1) con- 
stants of integration, making 2n-{- 6 unknown constants 
to be determined, as Jij, M^, R^^ and R^ are all unknown. 
Noting that R^ + R^ equals the sum of the loads, and 
that the moments of all the forces must be zero will give 

two, the values of y and -^ being zero at each wall will 

(XJC ~m 

give four more, the derived values for y and — in the 

(XX 

adjacent expressions for M are equal for values of x 
under any concentrated load will supply 2 n equations, 
the required 2n-|-6 equations may be written. If the 
loads are symmetrical with the middle of the beam, the 

solution will be much simplified, as — ^ is zero at the 

dx 



THE ELASTIC CURVE 109 

middle of the beam, and R^ = R^ and M^ = M^. The 
methods given are general and will completely determine 
the bending moments or deflections for any restrained 
beams. 

Having fully determined the expression for the values 
of M and V for any restrained beam, the shear and 
moment diagrams may be drawn, and the maximum 
moments as well as the inflection points determined, as for 
simple beams. The value of x which makes the vertical 
shear equal zero gives the dangerous section for the beam, 

MO 

and the formula S= — — - with the value of M a maximum 

can be used for all investigations of the strength and 
safety, as well as the design of restrained beams. 

Art. 67. Continuous Beams. 

A continuous beam was defined as one having more 
than two supports. 
The supports are j QOOOOOnO^i^^OO 1 



assumed to be on the I ^ 

same level and rigid, 
and the section of 
the beam uniform. 



fx-^ Uz i^-x-> 



Ajk A T 

\y 2i ^-—iz—^ 

Hi 

Fig. 67 



R^ R3 



nTI 



T 



Since there are more than two reactions, their values 
cannot be determined by the laws of mechanical equi- 
librium. Let Fig. 67 represent any two intermediate 
spans of a continuous beam or girder, whose lengths are l-^ 
and l^^ and the loads are w-^ and iv<^ per linear unit, respec- 
tively. Let iV^, iV2' ^^^^ ^z be the unknown bending 
moments at the supports, whose reactions R^, R^^ and R^ 
are also unknown. 

The moments iV^ and iVg may be assumed to be caused 
by couples acting to the left of Rj^ and to the right of R^. 



110 MECHANICS OF MATERIALS 

Considering the left hand span with the origin at i?^, and 
X the distance of any section of the beam from i?^, 

M= N^ + R^x - ^-^ (1) 

and ^/^, = JSr. + R,x - ^. (2) 

dx^ 2 

Integrating twice and determining the two constants of 
integration from the known conditions that ^ = when 
X equals either or I, 

£I^ = ]}^x + ^^-'^-^-\-a (3) 

dx 2 6 

and ^1/=^ + ^-'-^+ C,x + C„ (4) 

(7„ equals zero and C. =-J-J- Li xi_, 

^ ^ ^ 24 2 6 

Substituting this value of 0^ in (3), 

^/^ = iV^ :r 4- ^^ _ ^-M? + !^ _ ^ _ ^?iZl rs'-v 

dx ^ 2 624 2 6*^"^ 

If we take the origin at R^, and consider the right hand 
span, remember that the sum of the moments of the forces 
acting to the right of any section is the same as the sum 
of the moments of those acting on the left of the same 
section but with the opposite sign, and let x be any dis- 
tance in that span from R^, then 

and -EI^^m^R.x-'^, (5) 

dx^ ^ ^ 2 ' ^ ^ 

the sign of the moment iVg being left unchanged, as its 
value and real sign are unknown. 



THE ELASTIC CURVE 111 

Integrating this expression twice and determining the 
constants of integration from the known condition that 
y = when x equals either or l^, we have 

dx ^ 2 6 2^ 2 G ^ ^ 

and 

ETtJ = 3-^ _[_ -^3-^ ^2^ _[ '^2^2 ^ iV3^2'^' -^3^2 "^ /'7\ 

^ 26 24 24 2 6*^^ 

The values of -^ in (3') and (6) are equal, for x = Lm (^') 
dx 

and x = l^\n (6), (3') reduces to 

\dxr='' 2^3 8 ^ ^ 

— Ei(^\ = ^^^ I ^^ ^^^ r6'^ 

\dxJ--^^ 2 3 8 ' ^ ^ 

From (1), iHf = JST^ when :r = Z^ 
whence M= JSr^ = ]V^-{- R^\ - '-^, 

and ^ _ -^2~-^i _|_ !^^ /3^ 

(-J ^ 

Similarly, from (5), 3/= iVg when a; = Zg 
and j^ _ - ^2 ~ -^3 _ !M2.^ /'9^ 

Equating (3'^ ^i^d (6') and substituting the values R^ 
and jRg from (8) and (9), and reducing, we have 

N^l, + 2 NSi + h) + N,l, = - "'/i' + ^iV , (10) 

which gives the relation between the unknown bending 
moments iV^, N^^ and iVg, and the uniform loads iv-^ and w^. 



112 MECHANICS OF MATERIALS 

One equation similar to (10) may be written for any three 
consecutive supports, and if n is the number of the sup- 
ports, n— 2 such equations may be written. 

When the ends of the beam are merely supported, the 
moments are iVj, JV^ . . . I^n^ and the values of iV^ and iV„ 
are each equal to zero, so that the n — 2 equations that 
may be written will completely determine the values of 
the ?i — 2 unknown moments. 

When both ends of the beam are to be fixed, the two 

dy 
additional conditions that — = at either end support, 

ax 

will furnish two more equations that will be sufficient to 

determine the unknown bending moments at the end 

supports. 

When the values of the bending moments at the supports 
are determined, the reactions at each support may be found 
by the use of equations similar to (8) and (9). 

When there are concentrated loads on each span, the 

discussion is complicated by the fact that there are two or 

more forms for the equation of the elastic curve for any 

span instead of only one, bringing into the problem two 

constants of integration for each concentrated load. 

di/ 
Noting the additional conditions that the values of -^ 

dx 

and ^ where two equations meet at the point of application 

of any concentrated load are equal for the value of x at 

that point, it is simply a question of algebra to find the 

relation between the three moments. 

The expression as given in equation (10) is known as 
the Theorem of Three Moments and was published in 1857. 

Knowing the reactions and the loads, the shear and 
moment diagrams may be drawn as for simple beams. 



THE ELASTIC CURVE 113 

The maximum bending and shearing stresses may be 

found by using the maximum values of M and V in the 

Mc V 

fundamental formulas S= -— and S=—. 

I A 

The maximum bending moments occur where the shear 
passes through zero; hence, by writing the expression for 
the vertical shear and equating it to zero, the position of 
the maximum moment may be found and its magnitude 
determined from the equation for the bending moment. 
Equating the expression for the bending moment to zero 
will give the point of inflection. The caution given before 
may well be repeated here : 

'' When suhstituting the value of x which renders the verti- 
cal shear zero to determine the value of the maximum bending 
moment^ substitute in the particular expression for the value 
of M that applies to that portion of the beam, and tvhen 
equating the expression for M to zero in order to find the in- 
flection points, the expressiofi which applies to that portion 
of the beam must be used.^'' 

As was the case with overhanging beams, the inflection 
points and the points of zero shear may be approximately 
located by inspection of the moment and shear diagrams. 

The maximum stress in a continuous beam with a num- 
ber of equal spans may be less than the maximum stress 
for simple beams for the same spans ; hence, when using 
continuous beams care must be taken to insure that all 
the supports are on the same level and practically rigid, 
otherwise the maximum stress may be greater than for 
a number of simple beams. Take the case of two equal 
spans with a uniform load of w lb. /ft. If the beam is 
truly continuous, that is the supports on the same level 
and practically rigid, the maximum unit stress is, — \ ivl^. 



114 MECHANICS OF MATERIALS 

This is numerically equal to the maximum moment in a 
simple beam of the same span. 

Suppose the middle support of the continuous beam to 
sink so that there is no reaction, then the maximum 
moment is ^ w 4:1^ = ^ wT?^ instead of \ wP, or the unit 
stress will be four times as large as for the simple beam. 

As the deflection of beams as used in engineering 
structures is a very small quantity, the stress may easily 
approximate this maximum value when the supports do 
not remain precisely at the same level. 

EXAMINATION 

What is meant by the expression, "the elastic curve of a 
beam"? 

Derive the differential equation for the elastic curve of 
any beam. 

Name all the conditions that must be satisfied if the 
use of the equation will give results that are approximately 
correct. 

What is the " deflection " of a beam ? 

Show how the differential equation of the elastic curve 
may be used to determine the maximum deflection for any 
beam. 

Prove that if the loading of any simple beam is symmet- 

rical with the middle of the beam, -e^ = at that point. 

ax 

Why is it that when there are n concentrated loads on 
a beam, the expression for the bending moment may take 
-^4-1 different forms ? 

Show that the maximum deflections of simple and canti- 
lever beams rectangular in section, and uniformly loaded, 

vary as — and —3. 

What is a restrained beam ? 



THE ELASTIC CURVE 115 

Why is it that the laws of mechanical equilibrium can 
not be used to determine the reactions for restrained 
beams? 

Explain how the differential equation of the elastic 
curve may be used as an aid in the determination of the 
reactions for restrained beams. 

Explain how to find the maximum bending moment 
for any restrained beam. 

Will the maximum bending moment and the maximum 
deflection always be found at the same point in the beam ? 

State under what conditions they will be found at the 
«ame point. 

What is a continuous beam ? 

State the various steps in the method used in deriving 
"the equation known as " the equation of three moments." 

Write " the three-moment equation " for a continuous 
beam with equal spans and a uniform load on each span. 

Show how to find the reactions for a continuous beam 
after the equation of three moments has been found. 

Show that in the case of a continuous beam uniformly 
loaded the maximum unit stress may accidentally be greater 
than the maximum unit stress for a number of simple 
beams, one for each span. 

When the size of a beam for a given span has been 
found on the assumption that the beam was to be fixed at 
l)oth ends, it is necessary to take precautions to preserve 
the restrainment. Why ? 

Show that in any uniformly loaded beam, simple, 
restrained, or continuous, if l^^is the vertical shear on the 
right of any left hand support, and JV is the bending 
moment at that support, the maximum bending moment 

T^ 

in that span is ilSfJ^iax. = iV4- - — , in which w is the uniform 

load per inch. 



116 MECHANICS OF MATERIALS 



PROBLEMS 

1. Show that if the depth of a rectangular beam of 
uniform strength is constant, the elastic curve is a circle. 

S c • • 

SoLUTiox. We have the relation _ = _, from which r is constant 

E r 

when c is constant, but c=-; hence, r is constant and the curve a. 
circle. 

2. Show that if a simple beam carries two equal loads 
at equal distances from either end of the beam, the elastic 
curve between the loads is a circle. 

3. A simple beam BO ft. long carries a uniform load 
of 160 lb. /ft. The beam is 4 in. wide and 6 in. deep. 
Modulus of elasticity 1,600,000. Required, the inclina- 
tion of the beam with the horizontal at the supports. 

Solution, Find the value of -^ from the differential equation 

dx 

of the elastic curve ; put x = and solve. 

4. Find the maximum deflection for a simple beam, 
length Z, modulus of elasticity E, moment of inertia Z, 
loaded with 

(^) a uniform load of iv lb. /in. 

(6) a single load P at the middle. 

Solution for (a). The expression for the bending moment 

at any section of the beam is M = R^x — and B^ = ^', hence 

EI — - — is the differential equation of the elastic curve for 

this beam. 
Integrating, 

£:/ ^ = '^ - ^' + C; but^ = when x = K therefore C = - ^'; 
f/x 4 6 dx 2 24' 

1 u i'i i- -i. 1 -m dii idx- wx^ id^ 
and substitutms: its value. El ^^ = . 

dx 4 6 24 

Integrating again, 

£:/y = !^^ _ EZ! _ !^ + C, ; but as V = when x = 0, C, = 0, and 
12 24 24 ' 1 ' 

the equation of the elastic curve becomes 24 Ely = 2 idx^ — wx"^ —wl^x^ 



THE ELASTIC CURVE 117 

Equating the expression for -^ to zero, we find that x = --, — .365/, 

dx 2 

and 1.365 L The latter values have no meaning for this problem, and 
?/ is a maximum when x = -. Substituting this value for x in the 
expression will give the value of the maximum deflection. 

5. Find the maximum deflection of a cantilever beam, 
length I inches, modulus of elasticity U^ moment of 
inertia /, loaded with 

(a) a uniform load of w lb. /in. 
(5) a single load at the end. 

Solve by taking the origin at the w^all and measuring x toward the 
free end, and also at the free end of the beam. 

((?) a single load P, at a distance kl from the wall. 
Suggestion for (c). For any section on the left of P, M = 0', 

hence that part of the beam is straight. The value of -^ for x = kl 

dx 

gives the slope of the straight portion. When the deflection under 
the load is known, the deflection of the end of the beam may easily 
be found. Take the origin at the wall. 

6. A steel cantilever beam rectangular in section is 
18 in. in length, and is to carry a load of 1000 lb. at 
the free end, and deflect .36 in. under that load. The 
unit stress may be taken as 30,000 Ib./sq. in. Find 
the size of the beam. 

7. A beam fixed at one end and supported at the other 
is I in. long and carries, 

(^a^ a uniform load of w lb. /in. 
(5) a single load P at the middle ; 
find the expressions for, 

(1) the deflection at any point of the beam. 

(2) the maximum deflection and the point where it 
occurs, 

(3) the maximum bending moment and where it occurs, 

(4) the reaction at the supported end of the beam, 
and make a sketch showing the form of the shear and 
moment diagrams. 



118 MECHANICS OF MATERIALS 

Solution for (a). Taking the origin at the supported end let 
R^ be the unknown reaction at the supported end. The value of 

M at any point of the beam is 31 = R^x , as there are no con- 
centrated loads on the beam, and the differential equation of the 
elastic curve becomes 

Integrating, £/ 1 = fe' _ !!|! + p. (2) 

The tangent at the fixed end is zero, hence -j-= when x = 1, and 

?r/3 R r^ 
substituting x=l \\\ (2) gives C = —; ^. Substitute this value 

of C in (2) and integrate, 

^^ R.x^ wx^ u'l^x RJ^x _, ,„. 

This equation must be true for x = when y = 0, therefore C^ = 0. 
It must also be true for y = and x = l; making y = and x — ly 
(3) becomes 

RJ^ wl^ id^ Rd^ ^ „ ?>id 

Substituting for Cj and R^ in (3), we have 

48 Ely = 3 idx^ - 2 wx^ - wl^x (3') 

as the equation of the elastic curve for this beam. 

The value of M was R.x , and as R, = , 

2 8 

T,^ 3 ivlx WX^ ,4x 

M=^-—, (4) 

also F = 7?j — lox — wx^ (5) 

8 

and from these two equations the shear and moment diagrams may 

be drawn. 

3 I 
F = for x — — ; therefore the maximum moment occurs at f /, 

8 

and this value of x substituted in (4) gives 

, ;. _ 9 wV^ 
' ~ "128"* 



THE ELASTIC CURVE 119 

There will be a negative moment at the wall, and its value can be 
obtained by making x = I in (4), or 

X-l g 

Making M = and solving (4) for x gives 

T, , 3 Wlx WX^ rx 3 ? 

M = -g ~ = 0,ovx = -, 

as the inflection point. 

Substituting for C and R^ in equation (2) gives 

j^jf^y _8 wlx^ vjx^ tcl^ o icl^ 
dx~~lQ tT "O l6~' 

This expression is the differential coefficient of (3) ; therefore, 
equating the right hand member to zero and solving for x gives the 
value of X for which (3) is a maximum. This results in a cubic 
equation, S x^ — 9 Ix"^ -{- l^ = 0. The roots of this equation are x = I, 
.42 I, and — .298 I. The latter value has no meaning for this beam, 
and x = l is a minimum; therefore the maximum deflection occurs 
at a; = .42 l. 

Substituting x, .42 I in (3) gives ymax. = .0054 Z as the maximum 
deflection. 

Suggestion for (b). There are two differential equations for 
the elastic curve and the two curves have a common tangent and 

common deflection for a value oi x = -. 

2 

8. A beam I in. long is fixed at both ends and carries, 
(a) a uniform load of w lb. /in., 
(5) a single load P at the middle ; 
find 

(1) the expression for the deflection at any point of 
the beam, 

(2) the expression for the maximum bending moment 
and its position, 

(3) the expression for the maximum deflection and 
where it occurs. 

(4) Make a sketch showing the form of the shear and 
moment diagrams. 



120 MECHANICS OF MATERIALS 

Suggestion for (b). Since the loads are symmetrical with the 

P dy 

middle of the beam M^ = M^ and R^ = R^ = — and -r^ = 0, when 

a: = l, X = -, and x — 0. 

2 

9. A rectangular beam, 20 ft. long fixed at both ends 
carries a uniform load of 100 lb. / ft. If the modulus of 
elasticity is 1,500,000, the safe unit stress is 500 Ib./sq. in. 
and c? = 4 5, find the size of the beam. 

10. If the beam in 9 carried a load of 2000 lb. at the 
middle, and the other data the same, find the size of the 
beam. 

11. Find the maximum deflections for the beams in 
problems 9 and 10. 

12. Draw the shear and moment diagrams for the 
beams as given in problems 9 and 10. 

13. If the beams in problems 9 and 10 were to be beams 
of uniform strength with a constant depth, sketch the 
plan and elevation for each beam. 

14. Sketch the shear and moment diagrams for the 
beams in problem 13, using as ordinates the unit shearing 
stresses instead of the vertical shears, and the unit bending 
stress instead of the bending moments. 

15. Select a standard steel I beam to be used as a con- 
tinuous girder for four equal spans of 8 ft. each. The 
-ends are simply supported and the beam is to carry a uni- 
form load of 7000 lb. / ft. Unit stress not greater than 
16,000 1b. /sq. in. 

Solution. As the ends are supported, N^ = N^ = and from the 
symmetry of the spans and loads, iVg = N^ and i?j = R^, R^ = R^. 
The three-moment equation for equal loads and sj)ans reduces to 

N, + ^N, + N,= -'-f. (1) 

Beginning with the second span, 

N.-V^N. + N.^-"^, (2) 



THE ELASTIC CURVE 121 

and with the third span, 

N, + ^N, + N,= -'-f. (3) 

Since iVg = N^, (2) becomes 

2N, + iN, = - 'J^, (4) 

and as iV. = 0, eliminating N^ from (3) and (4), we find that 

N = - — 

Making N■^^ = and substituting for N.^ in (1), 

^ _ _ 3 wP 

' ~ 28 ■ 

Substitute the value of lYg i^^ equation (8) of the text, and we have 

p p 11 wZ 

Ti, ilr = . 

^ ^ 28 

In the second span, if Vo is the vertical shear just on the right of 
R2, the shear at any point is F = Fg — wx, where x is the distance of 

Y 
the point from jRg* F is evidently zero when x = —^. 

The bending moment at any point of the span is 

Y2 

hence Fg = itj + 7?2 ~ ^^'^* 

Substituting this value of Fg in the expression for maximum mo- 

ment gives as the maximum bending moment in the second span — — . 
The maximum moment in the first span is 

R^x-'-^ = .077 wl\ N,= - ^, and N, = -j^, 

and as the moments in the third and fourth spans are the same as 
in the second and first, the maximum moment occurs at the second 

Q /2 Q TTT'/ 

support and is , which may be written , where W is the total 

load on one span. 

Substituting this value of M in —^ = -, we find the value of / 
as 36. ^ ^ 

From the tables we find that a 12 in. I beam weighing 31.5 lb. / ft. 
has a value of 36, and that beam will be chosen. 



122 MECHANICS OF MATERIALS 

16. I beams are to be chosen to cover two equal spans 
of 10 ft. each. The uniform load for each span is 5000 
lb. / ft. and the maximum allowable unit stress is 15,000 
lb. /sq. in. Choose a standard I beam on the assumption 
that the beam is continuous. . What beam must be used 
if two simple beams are used instead ? 

17. Select a standard steel I beam to cover three spans 
of 8 ft. each and carry a uniform load of 7000 lb. /ft. The 
unit stress is not to exceed 16,000 Ib./sq. in. What size 
beam would it be necessary to use if each span was covered 
by a simple beam ? How much weight of steel will the 
use of continuous beams save ? 

18. In problem 17 find the reactions at the supports for 
each kind of a beam. 

19. Select a continuous steel I beam to cover two equal 
spans of 12 ft. each, and carry a load of 36,000 lb. at the 
middle of each span. Unit stress 12,000 Ib./sq. in. 

Suggestion. There are four forms of the differential equation 
of the elastic curve, each two having a common tangent and deflection 
at the load or reaction where they meet. The three reactions are all 
unknown but R^ = R^, and there is an unknown bending moment 
at i?25 the middle reaction. 



CHAPTER VI 
LONG COLUMNS 

Art. 68. Stresses in Long Columns. 

A theoretical bar under the action of axial compression 
should always yield by crushing, since the resultant of 
the stresses and the axial loads are in the same line. In 
dealing with the actual materials and loads, the force 
due to the load is not always axial, no matter how much 
care is taken to make it so, and no material in common 
use is absolutely uniform. In nearly every case of bars 
under axial compression, there is more or less of a couple 
caused by the bar not being straight and uniform in struc- 
ture, or the loads not axially applied. 

The effect of this couple is to produce bending in the bar. 
As the amount of bending or deflection of a beam under 
the action of bending forces varies directly with the cube 
of the length, it is easy to see that the danger of bending 
a bar under the action of compressive forces which are not 
exactly axial, increases very rapidly with the length, and 
also that any bending increases the moment of the axial 
forces. 

Experiment has proven that when the length of the bar 
does not exceed ten times its least dimension, it is more 
•liable to fail by crushing, than by bending and crushing 
combined. 

123 



124 MECHANICS OF MATERIALS 

Therefore, when the length of a bar under axial com- 
pression does not exceed ten times the least dimension, 
the formula P = AS may be used to determine the rela- 
tions between the load and unit stresses. 

Such a bar is called a short column or strut. When the 
length of the bar exceeds this limit, it is called a long 
column; and as the stress which is a combination of bend- 
ing and comj)ressive stresses is not uniformly distributed 
over the area of the cross section, the formula P =AS no 
longer applies. 

Since the amount of the bending stress in a long column 
is due entirely to the imperfections of the material and the 
eccentricity of the loads, there can be no strictly theoretical 
formula developed for the determination of such stresses. 

There are two formulas in general use expressing the 
relation between the load and the dimensions of a long 
column: Euler's formula for long columns, which is de- 
rived from theoretical considerations, gives the relation 
between the load and the elastic resistance of the column, 
taking no account of the unit stress induced ; and Ran- 
kine's formula, which gives a relation between the loads 
and the unit stress induced, while theoretical in form, 
contains an empirical constant. 

Art. 69. Euler's Formula. 

This is the oldest discussion of the theory of long 
columns, and was offered by Euler in 1757. 

He assumed: 

1st, that the column was perfectl}^ straight, 

2d, that the loads were axially applied at the center of 
gravity of the cross section, 

3d, that the material was uniform in density. 



LONG COLUMNS 



125 



Such a column would never bend under any load, but 
would fail by crushing. To produce incipient bending he 
assumed a slight lateral force to be exerted against the 
column while it was under the axial load, and determined 
the value of the axial load P that would heep the column 
bent after the lateral force was removed. 

The elastic curve of a bent column may take any one of 
several forms, depending on the condition of the ends. 



Art. 70. Columns with " Round " 

If the ends of the column are de- 
signed so that there will be no restraint 
to the tendency to rotation about a 
point on the end, the column is said to 
have ^^ Mound " or "Pm " ends. If the 
ends are rounded as in Fig. 70, it will 
bend in a single curve, and the elastic 
curve may be represented as in the 
figure. 

Taking the origin at 0, the moment 
of -P about a point in any section dis- 
tant X from is M = Pa — Pi/, and 
the equation of the ehistic curve for 
this case is 

UI^=Pa- Pij. 

dx^ ^ 



or " Pin " Ends. 




(1) 



Multiplying by 2 dy and integrating with respect to y, 

'dy\^ 



dii 
Evidently -^ is when «/ = 0, hence C-^ = 0. 



(2) 



126 



MECHAXICS OF MATERIALS 



Extracting the square root and transposing, 



integrating, 



^ \ 1 ay — y^ 



X 



= \ 



P 



vers-i ^4- a 



a 



'2. 



(3) 



(4) 



When a: = 0, then ?/ = a and vers"^ — = ^' 

a 'Z 

hence ^ = --\/-p* 

Also when x = I, y = 0, and vers~i - = 0, 



hence 



1- '^■.1^^ 
2 ~ 2^ P ' 



or P = 






which gives the value of the load P that will keep a 
column with round ends bent, after the lateral force has 

been removed. As the deflection y has 
disappeared from the expression for P, 
its value is independent of the amount 
of bendinof. 



Art. 71. Columns with Square, Flat, 
or Fixed Ends. 
When the column is so designed 
that the tangent to the elastic curve 
at each end will be parallel to the 
length of the column, the column is 
said to have square, flat, or fixed ends. 
A column of this type may have a cap, 
which on account of the large surface 
it presents, tends to prevent free bend- 
ing about a point in that end, or it may 




Fig. 71. 



LONG COLUMNS 127 

have the ends firmly imbedded in masonry. The form 
that the elastic curve takes in this case is similar to that 
of a beam fixed at both ends when there is a single 
load at the middle of the span. Let Fig. 71 represent 
a column with fixed ends as bent by the load P. The 
moments that keep the ends of the column vertical being 
denoted by M^ and M^, the moment due to 31^ and the 
load P, at any section of the column distant x from O, is 

M= M^-\rPa-Py, 

and the equation of the elastic curve is 

EI ^=M^-\-Pa- Py. (5) 

Keducing as before and integrating, 

Eli^ = 2 M,y + 2Paij-Py^ Cg, (6) 

when «/ = 0, -^ = 0, and Q^ becomes zero, also when y = a^ 

-^ = 0, and the value of 3L is found to be • • 

dx ^ 2 

Inserting these values in (6), taking square roots, and 
transposing, 

^ Way — y^ 
Integrating (7), we have 

^ = ^/5vers-i2i/ + t7,. (8) 

^ P a 

Here x = when y = a\ 

hence vers"i ^ = tt, and C^ = — tt v — ; 

1 2 ?/ 
also when x = -^ V = ^i therefore vers~i -^= 0- 

2 "^ a 



128 



MECHANICS OF MATERIALS 



Inserting these values in (8), we have 

- = — TT \ , or i^ = 4 1 

2 ^ P V' 

which is Euler's formula for columns with fixed ends. 



Art. 72. Columns with Round and Square Ends. 

When one end of a column is restrained so that the 
tangent to the elastic curve at that end is always vertical, 

and the other end is left free to ro- 
tate about a point in that end, the 
column is said to be one with round 
and square ends. 

The elastic curve will take some 
such form as shown in Fig. 72. 
Taking the axis of X as vertical 
and the origin at the upper end of 
the column, if M^ is the moment that 
keeps the tangent vertical, the mo- 
ment due to M^ and that of the force 
P about a jDoint in any section of 
the column distant x from the upper 
end, is M^ M^-\- Pa — Py, and the 




equation of the elastic curve becomes 
EI-pi = M^-\-Pa-Py. 



(9) 



Equation (9) is the same form as (5), and as the conditions 



to be satisfied are the same with the exception that 
does not become zero when x equals -, we may write 



dx 



EI 
P 



..-1^-1/ 



^ ^ a ^ 



EI 

P ' 



(10) 



LONG COLUMNS 129 

Taking the origin at the lower end of the column, measur- 
ing X up from Oj, and remembering that the value of the 
bending moment for one end of the column is the same as 
that for the other end with the sign changed, we may 
write 72 

^^-i.=^^-^y' (11) 

This is of the same form as equation (1) ; therefore 

^[EI _, y IT [EI ..ON 

^=Vp-^ers ii__^_. (12) 

If h and d are two such numbers that - -f- - = Z, then 

d 

equations (10) and (12) are equal when x = - in (10) and 

I ^ 

a^= — in (12), and when they are equal, j/= 0. 
d 

Equating (10) and (12) and making ?/ = 0, we have 



(V 



EIit\, -.J.EI 1 .3 1 . q 

6 = d\ —— TT, whence o = - and d= 6» 

F 1) ^ P 2 



2 I 
Substituting x — ^ and ^ = in equation (10), we have 
3 

2? \~EI ^ ^EIiT^ 



= -7rV-p,orP=- 



3 ^ F 4: P ' 

as the formula for columns with round and square ends.* 

* The column is not in equilibrium for this case unless a couple H-H. 
(see Fig. 72) is introduced. Considering this couple and taking the 
origin at the round end of the column, the equation of the elastic curve 
becomes ^2,, 

which reduces to P = 2^^ (nearly), instead of - ^ J^ . The approxi- 
mate solution has been offered because the value of P as usually given is 

4 Z2 • 



130 MECHANICS OF MATERIALS 

In the solution of problems by the use of these formulas, 
1 is always the moment of inertia about a gravity axis 
perpendicular to the direction of bending. If the column is 
free to bend in any direction, the least value of I for that 
section must be used. E is the modulus of elasticity, and 
as the tabulated values are in inch pounds, P must be ex- 
pressed in pounds and I and Zin inches. It will be noticed 
that the formulas for the three types of long columns differ 
onl}^ by a constant. 

If the strength of a column is defined as the load it will 
carry, and the strength of a column with square ends is 
taken as unity, it is easily shown that the strengths are as: 

1 for a column with square ends, 

^ for a column with round ends, 

j9- for a column with round and square ends. 

Ar"^ niay be substituted for Zin the general formula for 
lonsf columns and the formula written 

P r'^Eir'^ 
—-= m , 

A l^ 

where m has values of 1, 4, |^, according to the condition of 
the ends of the columns, and r is the least radius of 
gyration. 

The ratio of the length to the least value of the radius 
of gyration is called the " slenderness ratio " of a column, 
and when this ratio is greater than 25, the column may be 
considered as a long column. 

If the stress was uniformly distributed over the section 

p 

of the column, the value of — would be the unit stress. 

P 

Takings — as the unit stress, m = 1, tt^ = 10, and the 

I ^ 
ratio of - = 100, the value of aS' for a wrought iron column 



LONG COLUMNS 131 

is 25,000 Ib./sq. in., which is practically the elastic limit 
of the material. 

If - is taken as 50, the value of jS becomes greater than 
r 

the ultimate strength of the material. 

As the majority of columns in engineering structures 

have the ratios of - from 50 to 150, it is plain that 

7' 

Euler's formula will not always give satisfactory results. 
The results of experimental work on long columns show 
that the formula can not be depended on unless the ratio 

of - is nearly 200. 

Euler's formula not being satisfactory for the usual 

range of the values of - has led to the adoption of a 

r 

formula, first proposed by Gordan and later modified by 
Rankine. 

Rankine's modification of Gordan's formula is in com- 
mon use among American engineers, while the European 
engineers prefer Euler's formula or some modification of it. 

If Euler's formula is used for the design of a long col- 
umn, the resulting dimensions should be checked by the 
formula for axial compression to determine the value of 
the unit stress. 

Art. 73. Rankine's Formula. 

This formula is based on the assumption that the maxi- 
mum unit stress in any section of a long column under 
axial compression is a combination of the compressive unit 
stress due to the axial compressive force and the maximum 
unit bending stress due to the probable moynent of the same 
force. 



132 MECHANICS OF MATERIALS 

Let Sc be the maximum unit compressive stress in any 

p 

section of a long column, S-^ = — be the unit compressive 

A 



stress due to the load P, S^ = ~- the maximum unit com- 



I 

pressive stress due to the probable moment of the axial 
force, and A the area of the section of the column; then 



S^=S^ + S^ ovS, = ^ + '^, (1) 

from which the maximum unit stress S^ may be found 

P Mc P 

when — ■ and -— are known. The value of — may easily 

Mc • . 

be found, but the value of —- is indeterminate, owing to 

the lack of knowledge of the probable eccentricity of the 
force P. 

If we assume that / is the maximum deflection due 
to the unknown moment M^ the value of that moment 
may be expressed as Pf. AYriting Ar^ for I and making 
M= Pf^ equation (1) reduces to 

'^. = f(l + $} (2) 

By analogy with beams where the maximum deflection 

12 . p 

varies as — ,/may be taken as proportional to -. If (^ 

c e 

is some number depending on the material of the column 
for its value, and w is a number whose value depends on 

the conditions at the ends of the column, then f = ncf) 
Substituting this value of / in equation (2), we have 

S, = ^fl + ncl>i\o.P=^^, (8) 



e 



LONG COLUMNS 133 

which is Rankine's formula for the solution of problems 
relating to long columns. Since S^ is usually given in 
pounds per square inch, P must be expressed in pounds, 
and Z and r in inches. The value of r to be used will 
•depend on the direction in which bending takes place. 
If there is no external restraint to bending, the least 
value of r for the section considered must be used. The 
value of (j) has to be determined by experimental work 
on long columns. The method generally taken to deter- 
mine the value of cj) is to find by experiment the load 
that will cause the column to fail, and substitute that 
value for P in Equation (3). S^ is taken as the ultimate 
compressive strength of the material and, as n, Z, JL, and 
r are all known, the value of (f) can be found. The relia- 
bility of the formula depends on the accuracy with which 
the value of <f) is determined. The following table gives 
the usual value of (j) as found for columns where the ratio 

of - varied from 20 to 200, and therefore can be applied 

^ I . 

to problems where the ratio of - is within these limits. 

r 



For hard steel 


6— 1 • 

T* 2 0000' 


for mild steel 


9 = 3 o^o 0"0 » 


for wrought iron 


^— 36 00 ' 


for cast iron 


r ~ eioo" ' 


for timber 


y ~ 3'0 0"* 



The strengths of columns of the same dimensions with 
•square, round, and round and square ends were found to 
be in the ratio of 1, ^, and -^q respectively. 

The strengths of the columns also varied inversely as P. 

Let Zj, Zg, and Zg be the lengths of three columns having 
(equal section areas and ends square, round, and round and 



134 MECHANICS OF MATERIALS 

square respectively ; then 1^ = 2 I^, where each column has 

4Z 
the same strength. Under the same conditions, l^ = —^' 

AS . ^ 

Writing Rankine's formula P = ^, and making I equal 

41 

to Zj, 2 I^. and — ^, successively, there will result a numerical 

coefficient for the last term in the denominator of 1, 4, ^^-. 
In Equation (3) n was a number depending for its value 
on the conditions at the ends of the columns, hence n 
must equal 1 for columns with square ends, 4 for columns 
with round ends, and ^^- for columns with round and 
square ends. 

As this constant n depends on the condition of the 
ends of the columns, and that condition determines the 
form that the elastic curve assumes, it is evident that n 
should bear some relation to the deflections of beams 
having similar elastic curves. 

Compare the maximum deflection of a simple beam 
carrying a concentrated load at tlie middle with the 
maximum deflection for a beam with both ends fixed 
carrying the same load. 

The elastic curve for the simple beam is similar to 
that for a column with round ends, and the deflection is 
four times as large as the deflection for the restrained 
beam, whose elastic curve resembles that for a column 
with both ends fixed. Therefore the value of n for a 
column with round ends should be four times that for 
a column with square ends. This agrees with the value 
derived from the assumptions of relative strengths. 



LONG COLUMNS 135 

Art. 74. Applications. 

In the use of Rankine's formula the value of aS'^ may be 
taken as the safe working unit stress, and then P will be 
the load that can be carried with safety. P is always less 

than aS'^, as l + ?^0 — is always greater than unity; hence, 

no matter how short a column may be, Rankine's formula 
will give a safe value for P when the safe value of S^ is 

used. Since n(f) is a small quantity, if the ratio of - is 

T 

small, 1 + ?z(/) — may be practically unity and the value of 

P very nearly equal to AS^* The value of <f) being 
derived from experiment, the formula is limited in its 

use to the range in the values of - covered by the exper- 

r 

imental work in its determination. The values of 0, as 
given, apply very well to all columns where the ratio of 

- lies between 20 and 200. Above this ratio the load 
r 

given by Rankine's formula will still be safe, as the 
formula gives the value of P too small. On the other 
hand, Euler's formula gives values for P that are too 

large when the ratio of - is less than 150, and the load 

r 

given does not agree well with practice till the ratio ap- 
proaches 200. 

The forces producing tensile, compressive, and bending 
stresses are all determinate, and the results given by the 
formulas for these stresses agree very well with those de- 
termined experimentally. The formula for the maximum 
unit stress in a long column contains the effect of an in- 
determinate bending moment, the actual stress depending 
on the magnitude of that moment. As no general assump- 



136 MECHAXICS OF MATERIALS 

tions can be made as to the probable value of the moment, 
the allowable working stress in long columns is always 
taken less than for the same materials under other forms 
of stress. Many other formulas, some of them entirely 
empirical and others more or less theoretical, have been 
proposed, but none have come into general use. For a 
full discussion of the various formulas the student is re- 
ferred to "Text-book on the Mechanics of Materials," 
Merriman. 

Cast iron columns are common in engineering work 
on account of the large compressive strength of the 
material. As long as the unit stress due to the bend- 
ing does not equal the unit compressive stress due to 
the load there is no tensile stress in the column. The 
columns are generally made hollow and round in section. 

Wrought iron and steel pipes are also often used for 
columns. The values of / or r may be found as soon 
as the internal and external diameters are known. 

Rolled steel shapes, in channel and I beams, are 
joined together by plates, which are riveted to them, 
and used as columns. Timber is used in the solid 
section and in the hollow box section. In the case of 
circular sections the value of I is the same for bending 
in all directions. When the column is " built " up, as is 
the case when the rolled steel shapes are used with tlie 
joining plates, the spacing should be such that the 
values of / for the " built " section will be equal about 
the two principal axes. The same is true of the wooden 
box sections. For the rolled steel shapes the value of 
I for the gravity axes of the rolled section, parallel and 
perpendicular to the web, are given in the tables, and 
must be transferred to parallel axes passing through the 



LONG COLUMNS 137 

center of gravity of the built section. For the sections 
where the elements are rectangular the value of the 

gravity moment of inertia is — —, where d is the dimension 
perpendicular to the axis. 

Although it involves considerable arithmetical and al- 
gebraic work, the process of determining the values of / 
for the "built" sections is simple. 

Given the moment of inertia of any area about an axis 
passing through the center of gravity of the area, as /^, the 
moment of inertia of the area about any parallel axis, dis- 
tant A, is / = /^ 4- Ah^. 

Having transferred all the gravity moments of inertia 
to a set of axes passing through the center of gravity of 
the built section, the sum of the moments with reference 
to either axis should be equal, and in case they are un- 
equal, the least value should be chosen for use in the long 
column formulas. 

EXAMINATION 

Explain why the formula P = AS does not give results 
that agree with experiment when the bar under axial com- 
pression is very long. 

Define a Long Column. 

What assumptions were made when the formula 

P = m — -— was derived ? 

What is meant by the expression " a column with round 
ends"? "a column with square ends" ? "a column with 
round and square ends " ? 

Derive Euler's formula for long columns with square 
ends ; with round ends. 

Explain why the formula as derived for columns with 
round and square ends is approximate. 



138 MECHANICS OF MATER^ALS 

Explain why the use of Euler's formula for long 
columns does not always give satisfactory results. 

Give the conditions under which the use of Euler's 
formula for long columns will give satisfactory results. 

State the assumptions that were made for the deriva- 
tion of Rankine's formula for long columns. 

Derive Rankine's formula for the strength of long 
columns. 

State the meaning of each symbol and the units to be 
used in making substitutions in the following formulas : 

P Er^iT^ ^ P S. 

and — r = 



A Z2 ^ 72 

1 + ^^*;:^ 

Does Rankine's formula for long columns give results 
that are more reliable than Euler's ? Why ? 

On what does the reliability of Rankine's formula 

depend ? 

Me 
The formula S = — - was used in the derivation of 

Rankine's formula for long columns. Does this fact limit 

the use of the formula to materials which satisfy the con- 

Mc 

ditions for the use of the formula S = — -? 

Show by the use of Euler's formulas that the strength 
of a column with square ends being taken as unity, the 
strength of a column of the same size with round ends is 
J and that of a column with round and square ends is -^^. 
Explain why the value of n in Rankine's formula is 
1 for a column with square ends 
4 for a column with rounds ends, and 
^^- for a column with round and square ends. 
Show that Rankine's formula applied to the solution for 
any column, no matter how short, will always give a safe load 
for that column. Is the same true for Euler's formula ? 



LONG COLUMNS 139 

Is there any limit to the length of a column to which 
Rankine's formula will apply? 

Why is the ratio of - used as limiting the use of 
either formula ? 

If the section of a column is a rectangle having one 
side 4 in. and the other 6 in., find the values of I and r 
to be used in the formulas for the strength of long 
columns. 

Why will a hollow cylindrical form make a stronger 
column than a solid cylinder of the same section area ? 

Given the moment of inertia about an axis through the 
center of gravity, how can you find the moment of inertia, 
of the section about an axis parallel to the gravity axis ? 

PROBLEMS 

1. A wooden column 10 ft. long is rectangular in sec- 
tion and has round ends. The sides of the rectangle are 
6 and 8 in. respectively. What is the maximum load the 
column will carry? 

Find the safe load. (Use Euler's formula.) 

Solution. The formula is P = ?I^, I=^-^=^2^=UX 

^' 12 12 

since the least moment of inertia must be used. 

E may be taken as 1,500,000 and tt^ used as 10. I = 120 in. 

Substitutino', 

P = 1.500,000 X U4 X 10 ^ ^ ^^^ 

14,400 

This is the greatest load that can be carried without failure by 

bending. The corresponding unit stress due to the axial force P is 

— = /" — = 3130 Ib./sq. in., or a stress about one half the ultimate 
yl 48 

strength of the material. In order to carry this load the column 
must be straight and the load axial. As these conditions are rarely 
ever satisfied, a factor of safety of at least 5 should be applied to the 
result. Using this factor, the safe load is 30,000 lb. and produces an 
axial unit stress of approximately 600 Ib./sq. in., which has a fair 
margin for safety. 



140 MECHANICS OF MATERIALS 

2. If the column in problem l was cast iron and had 
square ends, determine the maximum load it will carry 
and the unit stress induced by that load. Is it possible 
for the column to carry the load ? 

3. Show that a cast iron column with square ends must 

have the ratio of - approximately 90 in order that the 

r 

axial unit stress corresponding to the load P in Euler's 
formula shall be less than the ultimate compressive 
strength of the material. 

4. If the axial unit stress produced by a force equal to 
the value of P as derived by Euler's formula is equal to 

the elastic limit of mild steel, find the ratio of _ for the 

r 

column. Consider the ends round ; square ; and round 
and square. 

5. A standard 12 in. I beam weighing 35 lb. /ft. is 
to be used as a column with round ends. The length is 
10 ft. What load may be carried Avith a factor of safety 
of 5 ? If there should be a factor of 5 used with the 
formula for axial compression, is the load given by Euler's 
formula safe ? 

6. A cast iron column 20 ft. long has a hollow cir- 
cular section. The external diameter is 10 in. and 
the internal diameter is 8 in. Determine the value 
of the maximum load that can be carried, if the allowable 
unit stress for axial compression is 4000 Ib./sq. in. 
What will be the factor of safety against failure by bend- 
ing? (Column has square ends.) 

7. Solve problem 1 by the use of Rankine's formula. 



Solution. The formula isP = ^-t^-? -<I = 48 sq. in., <^ = .- 



I 144 
Sc = 10,000 lb. / sq. ill. for a maximum load, r^= — = - — = 3, / = 120 



LONG COLUMNS 



141 



■X 



«-3-^ 



<-3'- 



in. Therefore -^= 4800, and substituting these values, 
P= ^8x10,000 ^65 000 1^, 
1 + 3 0% 4800 

Comparing the results obtained from the solution by the two 
formulas, we see that the allowable load from the former is about 
twice that obtained by the use of Rankine's formula. Therefore a 
factor of safety of 5 on Rankine's formula is equal to using a factor 
of 10 with Euler's formula. 

8. Find the size of a circular wooden column 12 ft. 
long to carry a load of 50 tons with a unit stress of 
1200 lb. /sq. in. Column has square and round ends. 

9. A wooden column 15 ft. long 
has a box section as shown in the 
figure. Find the value of x so 
that the column will carry the 
greatest load possible. 

10. If the safe unit stress for 
the column as given in problem ■ ~^» -,f 
9 is 1000 Ib./sq. in., find the safe Problem 9. 

load for the column. 

11. Two standard 12-in. channel beams are to be used 
for a column. The channels are to be placed as shown 

in the figure and 

I I joined by lattice 

work. If the mo- 
ment of inertia of 
the lattice work is 
neglected, find the 
distance between the 
channels so that the 

column will carry the largest load possible. The column 

is to have square ends. 

12. If the safe unit stress in the column given in 
problem 11 is 12,000 Ib./sq. in., find the safe load. 



«M 



_^L 




Problem 11. 



142 MECHANICS OF MATERIALS 

13. A cast iron column 20 ft. long has a hollow cir- 
cular section. The internal diameter is 8 in. and the 
external diameter is 10 in. Required the load that may 
be carried when the maximum unit stress in the column 
does not exceed 4000 Ib./sq. in. (the column has square 
ends). Compare the result with that obtained from 
problem 6. 

14. The connecting rod for a steam engine has pins at 
either end that are parallel to each other. When bend- 
ing tends to take place in a direction perpendicular to 
the axis of the pins the rod acts as a column with round 
ends. When the bending tends to take place in a plane 
through the axis of the pins the rod becomes a column 
with square ends. If the section of the rod is rectangular, 
find the relation between the depth and breadth, so that 
the column may be equally safe against bending in either 
direction. (Use Euler's formula.) 

15. A cast iron column 20 ft. long has a hollow circular 
section 12 in. outside diameter. The allowable unit stress 
is 10,000 Ib./sq. in. and the load is 50 tons. Required 
the inside diameter of the column. Consider the column 
to have square ends. 

16. A standard 12-in. I beam 40 ft. long has braces 
along the web that prevent bending in a direction perpen- 
dicular to the web. If S^ is 12,000 Ib./sq. in., find the 
safe load for the column. Column has square ends. 



CHAPTER VII 
COMBINED STRESSES 

Art. 75. Stresses due to Force. 

When a force acts on any material, more than one 
kind of stress may be produced in any fiber of the mate- 
rial. In the previous chapters it was assumed that only 
one kind of stress resulted from the application of the 
force, and the magnitude of the stress calculated on that 
assumption. While this latter condition may be true 
in some cases, the force acting on a bar often produces 
two or more kinds of stress. These stresses when com- 
bined may result in a maximum unit stress greater than 
either of the original unit stresses. Any fiber of a beam 
has a tensile or compressive unit stress due to the bend- 
ing moment, and a unit shearing stress due to the ver- 
tical shear. A shaft carrying a pulley between two 
supporting bearings is a beam and a torsion bar combined. 
Each fiber has a unit stress due to the bending, a unit 
shearing stress due to the vertical shear, and a unit 
shearing stress due to the twisting moment transmitted 
through the shaft. 

Art. 76. Tension or Compression combined with Bending. 

When a bar is subjected to a force whose line of action 

is parallel to the axis of the bar, it was shown that this 

force could be replaced by an axial force and a couple. 

143 



144 MECHANICS OF MATERIALS 

The axial force produces either tension or compression, 
and the couple both tension and compression. In such 
a case, an approximate solution for the problem of finding 
the value of the combined stress is obtained by com- 
bining the tension or compression resulting from the 
couple, and the tension or compression due to the axial 
force. Let /S'j be the maximum unit stress due to both 
sets of force, S be the unit tensile or compressive stress 
resulting from the axial force, and Sf, the tensile or com- 
pressive unit stress due to bending, then S^ = Sf, + S, 

Me P 

where S^, = — - and 'S = — are both tensile or compressive 

stresses. 

When the axial force is a tensile force, the maximum 
stress found in this way is too large, as the axial force 
on the bent beam produces a moment Pj/ that tends to 
reduce the bending moment of the flexural forces. See 
Fig. 76. When the axial force produces compression, 
the moment Py of the axial force tends to increase the 
bending of the bar, and the approximate solution gives 
the resultant unit stress too small. As the error is due 
to the moment P^, when the deflection is small the 
error is also small. While the engineer as a rule desires 
to be as nearly accurate in his calculations as possible, 
when the approximate formula is simple and errs on the 
side of safety, it is often used in preference to the more 
exact and complicated formula. A bar is often designed 
to resist a combination of compression and bending stress 
by the use of the approximate formula, but when this 
is done the resulting dimensions should be used in a more 
exact expression and the actual unit stress determined. 

Let Fig. 76 represent a beam under the action of 
flexural and axial forces. Let 31 be the maximum mo- 



COMBINED STRESSES 



145 





Fig. 76. 



ment of the flex- 
ural forces, P an 
axial force which 
may be either 
tensile or com- 
pressive, /j the 
maximum deflec- 
tion of the beam. 
If M-^ and S-^ are the maximum bending moment and 
unit stress due to the sum of the moments M and Pfj, 

then S^ = — ^ = ^^ '^'^^ , the positive sign being used 

when P is a compressive force, and the negative sign when 
P produces a tensile stress. By analogy with beams 
under the action of flexural forces only, the value of 



/j may be assumed to be f^ 



^Ec 



, and the substitu- 



tion of this value in the equation for S^ gives 



S,= 



Mc a S^FP 
I ^ (S EI 



or S-^ = 



Mc 



I± 



aPl^ 
/8 E 



P 



The maximum unit stress is S. -\- —-= S\ where S. has 

^ A ^ 

the value as just found. S is either tension or compres- 

P 

sion, depending on the kind of stress represented by —• 

The values of a and ^ for various kinds of beam and 
loadings are given in the appendix. There are no values 
that apply strictly to this case, as the bending moment 
is increased by the value of the moment Pf^ Since 
/j is a small quantity, the error made in assuming that 
the values of a and /3 are those found for beams under 



146 MECHANICS OF IvfATERIALS 

flexural forces only is very small, hence a and y8 will be 

determined by the kind of a beam and the nature of the 

loading. The unit stress due to the effect of the two 

moments is jy£^ 

S^ = - 



i± 



while that due to the flexural forces only is Sf, — —- 
The difference is -pn^ 

if I ■ -^^'^ " 



^ E^ 

This difference divided by the value of S-^ and multi- 
plied by 100 is the percentage error in the bending unit 
stress when S^, is used instead of S-^. The value of this 

error is ± -——- 100, from which the per cent of error may 
EI 13 ^ -^ 

be easily found. As the amount of the error depends 

directly on the value of P, and inversely on E, Avhen P 

is small, and E large, the error will be so small that it 

may be neglected. The error also varies directly as P; 

therefore it is more liable to be serious when I is large. 

For a timber beam 20 ft. long, 6 in. wide, and 12 in. 
deep, carrying a load of 8000 lb. at the middle and at 
the same time a compressive load of 45,000 lb., the use 
of the approximate formula will result in an error of 
approximately 8 per cent. 

For a steel I beam, 10 ft. long, having a moment of 
inertia of 122, a concentrated load of 6000 lb. at the 
middle and an axial compressive force of 20,000 lb., the 
values of jS^ and S(, differ by less than one per cent, an 
error too small to be taken into account. 



COMBINED STRESSES 



147 



Art. 77. Roof Rafters. 

The common roof rafter is an example of a beam under 
tixial and bending forces. As a part of a truss there is some 
compression in the rafter, and the weight of the roof and the 
probable snow load produce both bending and compression. 

Let Fig. 77 represent a roof rafter, length I in., carry- 
ing a uniform load of w lb. in. The rafter is in equi- 
librium under the horizontal forces H^ and H^, V act- 
ing vertically at the 
wall, and the uniform 
load. Taking mo- 
ments about a point at 
the foot of the rafter, 

H^ sin (^ = — cos (^, 

Li 

from which 

The bending mo- 
ment of the force 

H^ about a point in a section distant x from the upper end 
of the rafter is H^x sin 0, while that for the uniform loads 




Fig. 77. 



on the left of the same section is 
stress due to the bending forces is 



2 



cos <^. The unit 



o _ Mc _ (^Hx sin ^ —\ icx^ cos <^) c 

\-Y J • 

The compressive force on the same section is ff^ cos (p due to 



wx 



the force If,, and — sin <b, due to the uniform loads on the 

section. Hence the total compressive stress in the section is 

P _ (ff cos <f> + wx sin (^) 



148 MECHANICS OF^IATERIALS 

Using the approximate formula, the maximum unit 
stress is 

^^, P , ri (^cos (f) + z^'x sin <^) (Hxsind) — I wx^ cos (j))c 
•^=1 + '^' = A + 1 • 

— cot (j) may be substituted for IT and the value of x that 

Li 

makes S a maximum found. Using this value of x^ the 
value of the maximum unit stress in the rafter may be 
calculated. If the rafter carries a single concentrated load 
at the middle, the maximum compression and bending 
stresses occur at the middle and the maximum stress is 
easily found. 

When there are several concentrated loads on the rafter,, 
the greatest unit stress will have to be found by trial. 
The compression between any two loads is constant; there- 
fore, the maximum unit stress between any two loads will 
occur at the point of maximum bending moment between 
tlie loads. The maximum stress in several sections may 
have to be found to determine the greatest unit stress in 
the rafter. 

Art. 78. Eccentric Axial Loads. 

When a beam is to resist axial as well as flexural forces,, 
it is nearly always possible to make the point of applica- 
tion of the axial force so that the moment about a point in 
the neutral axis of the mid-section of the beam will be 
equal and opposite to the moment of the flexural forces. 
If P (Fig. 78) is an eccentric axial force, and y is the dis- 
tance of its point of 
application above or 
below the neutral 
plane, then before 
any bending takes- 



^ 


i";: 


H^ 


-^ 


-.-€ 


-e- 




j\. 


Fig. 78. 


> ^ 





COMBINED STRESSES 



149 



place Py is the moment of the axial force about a point in 
the center of the beam. Let M be the maximum moment 
of the flexural forces ; then if the axial force is compression 
and the distance y measured below the neutral plane has 
such a value that M= Py^ the resultant bending moment 
will be zero. Similarly, when the axial force is tension, 
if y is measured above the neutral surface and its value 
taken so that Py = iHf, the resultant bending moment will 
also be zero. 



T<r- 



Art. 79. Shear and Axial Stress. 

When a bar which is subjected to an axial stress is 
acted on by forces at right angles with the axis, there are 
tensile or compressive and shearing stresses at every point 
in that bar. Let Fig. 79 represent an elementary cube 
cut from any 
portion of the A^V 

bar at which 
there are tensile 
or compressive 
stresses par- 
allel, and shear- 
ing stresses 
perpendicular to 

the axis. The tensile or compressive forces T^ and T^^ act 
on opposite sides of the cube, and the shearing forces Fj 
and T^ on parallel faces. Since they differ only by 
differential quantity, T^ = T^ and J\ = T^. The cube is 
not in equilibrium unless a pair of equal shearing forces 
^j, H^ are introduced. As the cube is in equilibrium and 
the arms of the couples are equal, it follows that V= H. 
Since the elementary block was a cube, the unit stresses 




Fig. 79. 



150 



MECHANICS OF MATERIALS 



due to the equal forces H and V must be equal. Hence 
at every point of the bar there exists a pair of equal 
unit shearing stresses at right angles to each other, in 
addition to the unit tensile or compressive stresses. The 
tensile or compressive and shearing unit stresses that exist 
at every point in the bar combine and create shearing and 
tensile or compressive unit stresses that are greater than 
the original unit stresses. 

Akt. 80. Maximum Stresses. 

To determine these maximum stresses let Fig. 80 repre- 
sent an elementary parallelepiped cut from any portion 
of the bar. 

Let its length be dx^ height dy, width unity, and the 
faces be parallel and perpendicular to the axis of the bar. 

The area on which the 
tensile or compressive 
forces act is dx times 
unity, while that on 
which the shearing forces 
act, is either dy or dx 
times unity. The forces 
that act on opposite sides 
may be considered to be 
equal since they differ by an infinitesimal quantity. Let 
aS'^ be the unit shearing stress and S the unit tensile or 
compressive stress. The force that acts perpendicular to 
the dy face is Sdy and the shearing force in the same 
plane is S^dy. The shearing force in the dx face is S^dx. 
Let ds be the diagonal, (^ the angle between dx and dz^ S„ 
the unit stress perpendicular to dz, and Sp the unit shear- 
ing stress along dz. Resolving the forces that act on 




Fig. 80. 



COMBINED STRESSES 151 

either side of dz into components parallel and perpendicu- 
lar to dz we have 

Spdz = S dy cos (f) + S^ dx cos (/> — S^ dy sin 0, (1) 

Sn dz = Sg dx sin (f) + S dy sin </> + Sg dy cos (f>. (2) 

Divide each of these equations by dz^ make -^ = sin cj) and 

— = COS (^, and equations (1) and (2) reduce to 
dz 

Sj^ = S sin <f) cos (fi-h Sg (cos^ <^ — sin^ c^), (3) 

Sj^ = S sin^ (^ + 2 aS'^ sin (f) cos (^. (4) 

Writing the equivalents of sin </> and cos (j) in terms of 

2 </), we have q 

S^ = ^ sin 2 (l>+S, cos 2 </>, (5) 

^^ = I (1 _ cos 2 (/)) + aS' sin 2 0. (6) 

Li 

By the usual process 

aS'^ is a maximum when tan 2 (f)= —^ ' 



jSp is a maximum when tan 2 ^ = 



aS' 

2S' 



Substituting these values of cf) in equations (5) and (6), 
laximum values of /S^ a] 



we find that the maximum values of jS^^ and S^ are 



In these expressions S,^ and aS' may be either tension or 
compression. When the positive sign is used before the 
radical, aS'^ is the same kind of stress as aS'. When the 
value of S^ is obtained by using the negative sign before 
the radical, aS',^ is compression when aS' is tension, and vice 
versa. When there is no shearing stress, aS'^ = and the 



152 MECHANICS OF MATERIALS 

value of aS'^ is S, while the value of S^ is ^ S. For Ss= 
the tan 2 (/> = oo or (/> = 45°. </> is the angle between the 
directions of the tensile or compressive unit stresses and 
the plane on which the maximum stress acts, hence the 
maximum S^, makes an angle of 45° or 135° with the direc- 
tion of aS'. Tlie angle that S^ makes with S under the 

same conditions is the tan"^ <^ = 0. This latter expression 

P . . 

shows that S = — will s^ive the maximum unit tensile or 
A 

compressive stress in a bar on which there are axial forces 
■only. 

These formulas are general and apply to all combina- 
tions of tensile and compressive with shearing stresses 
without regard to the nature of the force producing the 
stresses. 

When the unit stress S is induced by axial forces, the 
p 
value of aS' = — -. The stress S may also be due to a bend- 

ing moment, and in that case the value of aS' is taken from 
S = -— . aS'^ may be due to a simple shearing force or the 

result of a twisting moment. 

In the former case the value of aS'^ to be used is derived 

P PVG 

from aS', = — , and in the latter case aS', = — ^. When 
A J 

the bar under axial forces is a long column, Rankin e's 

formula for long columns must be used to find the value 

P f P\ 

of aS'. This formula may be written aS' = — f 1+ m(f)—U 

from which the value of aS' may be easily found. 

Art. 81. Horizontal Shear in Beams. 

There is a tensile or compressive unit stress in every 
fiber of a beam that is equal to aS' = — - and at the same 



COMBINED STRESSES 



153 




time a unit shear- 
ing stress resulting 
from the vertical 
shear. When the 
formula V=ASg_ 
was derived, it was 
assumed that the 
shearing stresses 
were uniformly 
distributed over 
the section of the 
beam. In the pre- 
vious article it was 
shown that there was a pair of shearing stresses at right 
angles to each other. 

Therefore in any beam there is also a horizontal unit 
shearing that is equal to the vertical unit shearing stress 
at all points of the beam. To deduce an expression for 
the horizontal unit shearing stress at any point of a beam 
imagine a parallelopipedon cut from the top half of any 
beam (Fig. 81). 

Let the length be dx^ the width 5, the distance of the 
lower side from the neutral axis ^j, and the distance of the 
top of the beam from the same axis c. The faces are to 
be taken as parallel and perpendicular to the axis of the 
beam. 

There are compressive forces acting on each end of the 
elementary block which vary in intensity directly as their 
distance from the neutral axis. Let aS' be the unit com- 
pressive stress at the upper surface, and dA a differential 

area at any distance 7/ from the neutral axis; then — i/dA 



154 MECHANICS OF MATEKTALS 

is the force acting at the distance y from the neutral axis. 
The sum of the horizontal forces acting on either end of the 
elementary block is the sum of the compressive stresses 

acting on the same area. Calling this sum H— — I ydA^ 

MS M C ^ ^^ 

and writing — for — , we have H = —- \ ydA. Let the 

I e I'^yy 

bending moments at the ends of the block be M^ and 
M2, and the sum of the compressive stresses on the same 
ends be H^ and H^. Since the ends are separated by 

dx, M^-M^= dM; then H^- H^ = ^C ydA. For 

I ^y^ 

equilibrium a force equal to H^ — H^ must be introduced, 

and this force must be equal to the sum of the shearing 

stresses on the area dx times h. 

This sum is a horizontal shearing force, and the unit 

stress is — \- ^. Therefore if Sf^ is the horizontal unit 

dM C 
shearing stress, Sj^ — — — — - I ydA. From the theory 

dx lo^yi 

of beams Vdx = dM ov — — = V. Substituting j^for — ^ 

dx „ ^^ cix 

in the expression for Sf^^ we have aS'^ = — - J ydA as the 

value of the unit shearing stress at a distance ?/j from 
the neutral axis. 

The parallelopipedon could have been cut from the 
lower half of the beam where the stress is tension and the 
reasoning would be equally true. The formula gives 
the value of the unit shearing stress at a distance y^ from 
the axis, in a section for which Vi^ the vertical shear as 
defined in the chapter on beams. The width of the beam 
at a distance y-^ from the axis is 5, /is the moment of inertia 

of the entire section, while I ydA is the static moment 



COMBINED STRESSES 155 

of the area of the section lying above the distance y-^ from 
the neutral axis. If c^ is taken as the distance of the 
center of gravity of the area above y-^ from the neutral axis 

ydA = a^c^^ and Sf^ = —- a^c^. 

^1 10 

For a point yi= c from the axis, a^c-^ = 0, and therefore 
aS'^^ = at the distance e from the neutral axis. When a^ 
is the whole area above the axis, Sf^ will be a maximum for 
that section. The greatest value of SJ^ for the beam will 
be found at the neutral surface in the section where the 
vertical shear is a maximum, since the value of aS'^ varies 
directly with F! Similarly, there will be no horizontal 
unit shearing stress in the sections where T^= 0. It has 
been proven that for any point in a section of a beam 
there was a horizontal unit shearing stress that was equal 
to the vertical unit shearing stress at the same point. 
The expression just derived for Sf^ shows that the hori- 
zontal unit shearing stress is a variable quantity in any 
section of the beam, therefore the vertical unit shearing 
stress must also be variable. For a rectangular section, 
breadth 6, depth d, the value of aS'^ for any section is 

- -— instead of S^ = — -, showing the maximum horizontal 
2od od 

shearing unit stress, and therefore the maximum vertical 
unit shearing stress is 50 per cent greater than the assump- 
tion of uniform distribution of stress would indicate. In 
determining the conditions for the safety of a beam, if the 

Mg 

unit stress derived from the formula S = —-- is a safe 

stress, the beam will in most cases be safely loaded. 
When a beam is short and deep, the horizontal unit 
shearing stress along the neutral surface may exceed the 



156 MECHANICS OF MATERIALS 

safe unit shearing stress. This is especially true of 
timber beams on account of the low value of the ultimate 
shearing strength of timber along the grain. Hence the 
value of the shearing stresses should always be investi- 
gated, as no beam is known to be safely loaded until the 
unit stresses of all kinds have been determined and found 
safe. 

Art. 82. Maximum Stresses in Beams. 

In the general theory of beams as presented in Chap- 
ter III, the shearing stress due to the vertical shear was 
assumed to be uniformly distributed over the area of the 
section of the beam. That this assumption was not 
strictly true is evident from the equality of the variable 
value of the horizontal unit shearing stress with the unit 
vertical shearing stress in the same section. The value 
of jS,^ being a maximum or zero, as V is a maximum or 
zero, shows that aS'^^ is zero when M is a maximum, or that 
there are no shearing stresses in the section for which M 

Mc 

is a maximum. The value of S as derived from aS^ = —- 

will, therefore, be the true unit stress for any section 
where itf is a maximum bending moment for the beam. 
The shearing stresses in the fibers along the upper and 
lower sides of the beam are also zero, since a^c^ = 0. 
Hence the unit stresses in these fibers at the various sec- 
tions of the beam will be the value of S -ds, derived from 

-— , when M is the bending moment for the section con- 
sidered. 

The unit tensile or compressive stress being zero along 
the neutral surface, the unit stress along that axis is one 



COMBINED STRESSES 157 

of shear, and its value may be found from the expression 
for Sf^. For simple beams 31 is zero at the supports 
and the unit stress at all points of the section is simply 

With these exceptions the unit stress at all points in 
a beam is a combination of the tensile or compressive 
stresses with the shearing stresses. The value of the 
maximum unit stress at any point in the beam may be 
found from the expressions for the maximum values of 
aS'„ or Sp^ when S/i is substituted for aS^^. These maximum 
stresses make angles with the axis of the beam that 
depend for their value on the relative values of S and S^. 

The unit stress given by -— - is the true unit stress, when 

il[f is a maximum moment, and it is easy to see by an in- 
spection of the expressions for the maximum values of 
Sn and Sp that their values can rarely ever exceed those 

given by — — and — I ydA, When a beam is deep 

vertically and carries a concentrated load at the center, 
the value of V is constant for all sections up to the 
middle of the beam. Therefore it is possible in such a 
beam that the maximum value of S^ may exceed the 

3Ic 

maximum value of S as found from — . This is espe- 
cially true of beams having I sections, as the value of the 
static moment of the flanges is nearly as large as the 
moment of the whole area above the neutral axis. The 
value of Sf^ in the web just below the flange being very 
large, for a section just to the left of the load where Vis 
a maximum and S nearly so, the values of aS',^ and aS^^ may 
be greater than the maximum values of S and Sj^, 



158 MECHANICS OF MATERIALS 

The shearing stress S^ at any point of the beam is a 
stress that is inclined at angles that vary from 0° to 45° 
with the axis of the beam. When the beam has an I 
section and is deep, these shearing stresses have a tend- 
ency to cause the web to buckle. To resist this tendency 
vertical angle irons are often riveted to the webs. When 
a beam of an I section is composed of angle irons riveted 
to a web plate, making what is known as a plate girder, 
the force on the rivets joining the angles to the web can 
be found from the values of >S'„ and aS'^^, and the areas over 
which these stresses are distributed. 

While the theory of beams as presented in Chapter III 
was defective in its assumptions regarding the distribution 
of the shearing stress, and its neglect of the combined 
stress, this discussion shows that for the majority of beams 

the formula S= -— will give the maximum bending stress 

when the value of itf is a maximum. The formula may 
be used for the design of all beams and the dimensions 
checked for safety by determining the values of aS'^, aS^, 
and Sf^. 

EXAMINATION 

Give some examples of bars where the forces acting 
produce more than one kind of stress. 

When a bar under bending forces also has a tensile or 
compressive axial force acting on it, why does not the 
resulting unit stress always equal the sum of the unit 
stresses due to each force ? 

Develop the expression 

^ _ 3/(? 



i±-^ 



^ E 



COMBINED STKE:5SES 159 

where jS^ is the maximum unit stress due to both the axial 
and bending forces acting on any bar. 

What assumptions are made in the development of the 
formula that are not strictly true ? 

When a beam which carries bending loads also has to 
resist tensile or compressive loads, how can the resulting 
bending moment be made practically zero ? 

When a bar is subjected to both axial and shearing 
forces, show that at every point of the bar there is a pair 
of equal unit shearing stresses whose directions make 
right angles with each other. 

What is the effect of the combination of shearing with 
unit tensile stresses ? Is the effect any different if the 
shearing stress is combined with an equal stress in com- 
pression ? 

Show how to determine the value of the maximum unit 
stress when tensile or compressive stresses are combined 
with shearing stress ? 

A bar is being acted on by tensile or compressive forces 
applied in the line of the axis. Is there any shearing 
stress ? How may its value be determined ? 

If the maximum tensile or compressive unit stress is 
given by 

and the positive sign is used before the radical, what kind 
of stress is aS^„? 

What is meant by the expression " horizontal shear " in 
beams ? 

Deduce an expression for the horizontal unit shearing 
stress at any point of a beam. 

Under what conditions will the unit stress given by 



160 MECHANICS OF MATERIALS 

aS'= — be the maximum unit tensile or compressive stress 

in a beam ? 

Under what conditions may the horizontal shearing 
stress become the most dangerous stress in the beam ? 

A deep I beam is short and carries a load at the middle. 
Is it possible that at some point in the beam there may 
be a unit tensile or compressive stress greater than that 

Mc 

given by S = —=- when the value of 3Iis that of the maxi- 
mum bending moment for the beam? 

When a beam is rectangular in section, where will the 
maximum unit shearing stress be found? Is the same 
true for beams of other sections ? 

PROBLEMS 

1. A roof with two equal rafters has a span of 40 
ft. and a rise of 15 ft. If the weight of the rafter is 
neglected, determine the size of the rafters 6 in. wide 
when each rafter carries a uniform load of 50 lb. per foot. 

Assume the allowable unit stress as 600 lb. / sq. in. 

2. Assume the rafters in problem l to carry a load of 
1000 lb. at the middle, instead of the uniform load, and 
find the depth of the rafters for the same unit stress. 

3. A simple wooden beam, 30 ft. long, 12 in. deep, and 
4 in. wide, carries a single load of 320 lb. at the middle 
and an axial compressive load of 14,400 lb. Find the 
maximum unit stress in the beam, 

(rt) by the approximate method, 
(h) by the more exact method. 

4. A simple steel I beam 20 ft. long, 12 in. deep, 
weighing 35 lb. / ft., carries a uniformly distributed load 



COMBINED STRESSES 161 

of 500 lb. /ft. and sustains a compressive load of 40,000 
lb. Is it safe if a factor of safety of 5 is needed ? 

5. Find the points of application of the compressive 
loads in problem 4 so that the unit stress resulting from 
the effect of the loads will be as small as possible. 

6. A steel eye bar 30 ft. long has a section area of 
2 by 6 in. The eye bar is to be used horizontally, and 
carries a tensile load of 288,000 lb. Determine the dis- 
tance from the center of the eyes to the axis of the bar so 
that the resulting unit stress will be as small as possible. 

7. Determine the value and location of the maximum 
unit shearing stress in the beam as given in problem 3. 

8. If the point of application of the compressive forces 
in problem 3 is taken so as to neutralize the effect of 
the central load, determine the value of the maximum 
shearing unit stress. 

9. A simple steel I beam, 20 ft. long, 12 in. deep, 
weighing 35 lb. /ft., carries a concentrated load of 30,000 
lb. at the middle. Determine the maximum horizontal 
unit shearing stress. What is the value of the total 
maximum shearing force ? 

10. Given the same beam as in problem 9, carrying 
the same loads, find the difference between the maximum 
unit tensile stress and the stress given by the formula 

a Mo 

11. A horizontal steel shaft, 5 in. in diameter, is 20 
ft. between bearings, and carries a load of 1200 lb. at 
the center. It transmits 250 H.P. at 100 revs. /min. 
Required the maximum unit stress induced. 

12. A standard 2-in. steel bolt is screwed so as to 
cause a unit tensile stress of 10,000 lb. /sq. in. What 



162 MECHANICS OF MATERIALS 

shearing force may be resisted if the maximum unit 
stress is not to exceed 15,000 lb. /sq. in.? 

13. How many 1-in. standard steel bolts must be used 
to resist a shearing force of 300,000 lb. if the tensile stress 
due to screwing up is assumed to be 10,000 lb. /sq. in. 
and the allowable maximum unit stress is taken as 12,000 
lb. /sq. in.? 

14. An angle iron is bolted to the side of an I beam 
and supports another beam whose reaction at the sup- 
ported end is 50,000 lb. Select a number of 1-in. bolts 
to carry the load. Assume that the unit stress in each 
bolt due to screwing up is 12,000 Ib./sq. in., and the 
factor of safety is 4. 



CHAPTER VIII 



COMPOUND BARS AND BEAMS 

Art. 83. Definition. 

When a bar is composed of more than one kind of 
material it is sometimes termed a compound 
bar. The formulas derived in the previous 
chapters apply only to bars made of one 
material throughout. This chapter will be 
devoted to the investigation of a few of the 
simpler cases of stress in the compound bars. 

Art. 84. Compound Columns; alternate layers. 

A column or pier built with alternate layers 
of different materials, as in Fig. 84, will evi- 
dently carry only the load the weaker section 
will support. The unit stress in any section 
may be found when the area of the section is known, as 
each section has to support the entire load. 

When the modulus of elasticity of the material and the 
length of each section are known, then the amount of 
shortening for each section can be found as for simple 
bars. The total shortening of the entire column is the 
summation of the deformations of the different sections. 

Art. 85. Compound Columns ; longitudinal layers. 

When a column is composed of different materials 
arranged longitudinally, the column becomes a bundle 

lf)3 



Fig. 84. 



164 



MECHANICS OF MATERIALS 



of simple bars. Each bar does not carry a part of the 

total load that is proportional to its area as in the i^ve- 

vioLis case. All the bars have the 

same amount of deformation, and when 

^ ^ the unit stress is within the elastic limit, 

^ j^ -^ the unit stress in each bar must vary 

4 Xy y\ directly as the modulus of elasticity. 

In Fig. 85 let U^, E^, U^, be the 

moduli of elasticity of the three bars, 

i Z 3 and the areas be A^^ A^, A^^ and the 

common deformation and length, e and 

I; then. 



-i__ 



z 



k t« I 



^1 



_s^ 



E^ = 



Sd 



^ e 



P 
Fig. 85. 



Writing for S-^^ S^^ and S^^ their values 
in terms of the loads and areas, we have 






A^3 



a) 



As the sum of the partial loads must be equal to the total 
load, 

P = P, + P, + P,. (2) 

Equations (1) and (2) will suf- 
fice to fully determine the values 
of the partial loads Pj, P2' ^^^^ 
Pg, when the length, moduli of 
elasticity, and areas are known. 

The above discussion will ap- 
ply equally well to compound 
bars under tensile forces, as the ^^^' ^^"" 

stress in each case is the result of an axial force. 




COMPOUXD BARS AXD BEAMS 



165 



Art. 86. Compound Beams. 

When a beam is composed of two different materials, 
the load carried by the beam is divided between the 
two beams of the different materials in a similar manner 
as in the case of a compound bar under compressive 
forces (Fig. 86). The deflection for each part of the 
loeam. is the same, and if the deflection is expressed as 



•^ I3FI 



, then in the given case 

,and W=W^-\-W^. 



In the above expressions W is the total load carried by 
the beam and W^ and W^ are 
the partial loads carried by 
each material of the beam. 

The equations will suffice 
to determine the safety of a 
compound beam under any 
loading. 

To design a beam to be 
composed of wood and steel, 
the size and shape of either 
material may be assumed and 
the load it will carry with a 

safe unit stress calculated. Either TF^ or W^ being found, 
the other becomes known at once. 

The shape of the required part of the beam may be 
found from the equations resulting from the equality of 
the deflections. The solution is a tentative one, as the 
unit stresses must all be within the elastic limit for the 
•equations to hold true. 




Fig. 866. 



166 MECHANICS OF MATERIALS 

Art. 87. Reenforced Concrete Beams. 

Concrete, being a material strong in compression and 
rather weak in tension, is not suitable as a material for 
use in beams. To remedy the defect in tensile strength, 
steel rods are embedded in the concrete below the neutral 
axis. The steel supplies the tensile strength that is lack- 
ing in the concrete, and as the concrete is fireproof and 
non-corrosive, the combination of concrete and steel, 
known as reenforced concrete, is used extensively in all 
building operations. Steel I beams are often imbedded 
in concrete, but this is done mainly to protect the beam 
from corrosion. 

The common theory for the design of reenforced con- 
crete beams neglects the tensile strength of the concrete 
and considers the entire tensile load to be carried on the: 
steel reenforcement. 

Experiments made on concrete in compression do not 
agree as to the possible variation of the modulus of elas- 
ticity with the unit stress. The theory given here assumes 
that the value of the modulus of elasticity is constant for 
all stresses that are allowable in engineering structures. 
The so-called parabolic formulas are derived on the assump- 
tion that the value of .£' decreases with an increase of stress. 
With a fair value of the ratio of the areas of steel to 
concrete the two formulas give nearly the same results. 

Art. 88. Straight Line Formula. 

In the derivation of the straight line formula for the 
strength of reenforced beams the following assumptions 
are made : 

1. That the unit compressive stress in the concrete 
varies directly as the distance from the neutral axis. 



COMPOUND BARS AND BEAMS 



167 



2. That the unit tensile stress in the steel reenforce- 
ment is constant and depends on the distance of its center 
of gravity from the neutral axis. 

3. That the entire tensile stress is carried by the steel 
reenforcement, the tensile strength of the concrete being 
neglected. 

4. That the neutral axis is located so that the tensile 
resistance of the steel is equal to the total compressive 
resistance of the concrete. 




I v2j_ 



Fig. 88. 



5. Perfect adhesion between the steel and concrete. 

6. That the value of the modulus of elasticity of the 
concrete is constant. 

Let Fig. 88 represent a portion of a reenforced concrete 
beam, and X-X the location of the dangerous section. 



168 MECHANICS OF MATERIALS 

Notation 

d = the depth of the beam. 
b = the breadth of the beam. 

z = the ratio of the sectional area of the steel reen- 
forcement to the sectional area of the concrete 
above the center of gravity of the steel. 
S^ = the maximum unit compressive stress in the concrete. 
S( = the maximum unit tensile stress in the steel. 
U^ = the modulus of elasticity of the concrete. 
Ug = the modulus of elasticity of the steel. 

7^ 

y = the ratio — ^. 

d^ = the distance of the outer compression fiber in the 

concrete to the center of gravity of the steel. 
X = the ratio of the depth of the neutral axis to the 
depth of the center of gravity of the steel, both 
being measured from the outer fiber of the con- 
crete in compression. 
xd^ = the distance of the neutral axis from the outer fiber 

of the concrete in compression. 
Mr = the maximum moment of resistance. 
31= the maximum bending moment for the beam. 
e = the depth of concrete below the center of gravity 

of the steel. 
I = the length of the beam. 
Linear unit, the inch ; unit of weight, the pound. 

For equilibrium, the resultant of the compressive 
stresses in the concrete must be equal to the tensile stress 
in the steel, and the moment of the couple formed by 
these two forces must be equal to the bending moment 
for the beam. 



COMPOUND BARS AND BEAMS 169 

Since all the stresses are within the elastic limit, the 
deformation of the fibers of the concrete and steel are pro- 
portional to their distance from the neutral axis ; hence 

Solving for S^ and writing ^ for the ratio of U^ to U,, 

we have 

S, = S, / . (2) 

?/(l - x^ 

Solving (2) for x, we have 



X = 



1 + 



^ (-3) 

The area over which the compressive forces is dis- 
tributed being xd^h while that for the steel is zd^b, and 
since the total forces are equal, 

zd,bS, = '^S„ovzS, = ^. (4) 

Substituting for x its value taken from (3) and solving 
for z^ we have 

^ = — Tory oT- (^) 



2 ?^ 1 + 



Equation (5) gives the required value to the ratio z 
in order that the given values of Sc and Sf may be de- 
veloped. (It must be noticed that z depends for its value 
on the relative values of S^^ Sf^ E^, and E^-, and can not be 
assumed at will, if the full strength of the steel and con- 
crete is to be made available.) 



170 MECHANICS OF MATERIALS 



Eliminating S^ from (2) and (4), we have 



X/" X \ x^ 



2 =T7 



(6) 



•2\il-x)yJ 2(l-:.)y 
and solving for x, 

Equation (7) gives the ratio x in terms of ^ and z; 
hence the location of the neutral axis may be found when 
these ratios are known. 

The value of the resisting moment can be found by 
taking the moment of the tensile force in the steel reen- 
forcement about a point in the line of action of the result- 
ant of the compressive forces in the concrete; therefore, 



Mr = S,zdMd^ - -^J = S^zhdMl - Ij. (8) 

The value of Mj. may be found in terms of S^ in a 
similar manner to be 

M^=S,^(l-l). (9) 

Equating the maximum bending moment for a given 
beam to the value of M^ as given by either equations 
(8) or (9) will give a relation between the bending 
moment and the dimensions of the beam. The resulting 
equations may be used to investigate the safety of a 
given beam or to design a beam to carry a given load. 

When the ratio z is given, equation (5) shows that 

the ratio of — ^ must have a definite value. 

If the given value of z is larger than necessary, the 
full strength of the steel will not be utilized. The 
allowable value of S^ becomes the limiting factor, and its 



COMPOUND BARS AND BEAMS 171 

value may be assumed. This assumed value used in 
equation (5) will determine the value of the unit stress 
in the steel. 

On the other hand, when the value given for z is too 
small, the safe unit stress in the steel becomes the limit- 
ing condition and the full safe strength of the concrete 
can not be made available. In any case neither S^ or aS'^ 
can exceed the elastic limit of the material and should 
not exceed a safe working stress. 

The depth, g, of the concrete below the center of grav- 
ity of the steel does not enter into the formula, as the 
tensile strength of the concrete is neglected. Evidently,, 
to get the most value for the steel reenforcement it should 
be placed as close to the lower side of the beam as pos- 
sible. The assumption that the bond between the steel 
and the concrete is perfect requires that there shall be 
a reasonable thickness of concrete around the steel. 
Therefore the depth e must be determined by practice. 
The depth should never be less than one inch. 

The extensive use of reenforced concrete is of compara- 
tively recent date, and while there has been a great deal 
of experimental work done on reenforced concrete beams 
and columns, much more will have to be done before the 
theory for their design can be considered as good as that 
for beams of one material. There are so many variable 
conditions to be taken into account that various experi- 
menters have arrived at seemingly contradictory results. 

The formula as given here is the one in general use 
and appears to give reliable results. The allowable unit 
stresses are generally taken lower than the usual practice 
for the same materials under conditions where the effect 
of the load is better understood. 



172 MECHANICS OF MATERIALS 

The form of the steel reenforcement is a subject to 
Avhich experimenters have given a great deal of attention, 
the object being to find the form that will insure the 
best bond between the steel and the concrete. 

The value of the modulus of elasticity for concrete 
depends on the proportions of cement, sand, and broken 
stone in the concrete. The value of E ranges from 
4,000,000 to as low as 850,000, the lower value being for 
a cinder concrete and the upper value for a 1 : 1|- : 3 
mixture. A fair average value for E may be assumed 
to be 3,000,000. Taking the modulus of elasticity of 

steel as 30,000,000, the ratio of ^^ = 10. The allowable 

unit stress in steel may be taken as varying from 10,000 
to 12,000 Ib./sq. in. for steady loads and the correspond- 
ing value of the safe unit stress in the concrete ranges 

from 500 to 600 Ib./sq. in. The ratio of -^ then be- 

comes about 20. 

These values inserted in equation (5) give a value of z 
as slightly over .8 per cent. The usual values of z range 
from .75 per cent to 1.50 per cent. 

To determine the safety of a reenforced concrete beam 
carrying a given load, first find the limiting value of the 
unit stress in the steel or concrete. This being done, 
the equation resulting from placing the maximum bend- 
ing moment of the beam equal to the right hand member 
of either equation (8) or (9), as the unit stress in the 
steel or concrete is the limiting stress, will suffice to 
determine the value of that stress. 

To design a reenforced beam, z may be assumed and 
limiting stress found, or z may be calculated by the use 



COMPOUND BARS AND BEAMS 173 

of either equations (4) or (5), using the maximum allow- 
able values for jSf and S^. Either b or d^ may be assumed, 
and equating the maximum bending moment for the 
beam to the expression for the resisting moment will 
suffice to determine the other dimension. The total 
depth d is equal to d-^ + e, where e must be assumed 
empirically. 

If z is assumed, the equation for the resisting moment 
that depends for its value on the limiting stress must 
be used. 

EXAMINATION 

What is a compound bar ? 

When a compound bar is used as a short column, will 
the formula for axial compression always give the true 
maximum unit stress ? Explain your answer fully, giving 
reasons. 

When a beam is composed of more than one kind of 
material, name the conditions that are used to determine 
the part of the load carried by each material. 

What is meant by " reenforced concrete " ? 

In the development of the " straight line " formula for 
the strength of a reenforced concrete beam, certain assump- 
tions are made. What are they ? 

Can the tensile stress in the steel, the maximum com- 
pressive stress in the concrete, and the ratio of the section 
area of the steel to that of the concrete be assumed at will ? 

How many of the three can be assumed ? 

Equation (8) of Art. 88 is an expression for the resist- 
ing moment for a reenforced concrete beam. Explain why 
it is true. 

If the ratio z is too large, what can you say of the unit 
stress in the steel ? 



174 MECHANICS OF MATERIALS 

If z is too small, can the full strength of the concrete be 
made available ? Explain fully. 

PROBLEMS 

1. A wooden column, section area 36 sq. in., is used to 
support a floor. The load on the floor is to be increased 
and a hollow circular cast iron column 6 in. external and 
4 in. internal diameter is placed beside the wooden column. 
What part of the total load will each carry ? 

2. A load of 60,000 lb. is to be carried on a hollow cast 
iron column 6 in. internal diameter. The interior of the 
column is filled with concrete ; how thick should the cast 
iron be if the maximum unit stresses in the cast iron and 
concrete are 3000 and 600 Ib./sq. in., respectively? 

3. If the column in problem 2 was filled with wood 
instead of concrete, how thick should the cast iron be 
made ? 

4. A standard 6-in. steel pipe is encased in concrete 
2 in. thick and used as a column. How much more load 
may be carried with, than without, the concrete ? 

5. Two standard 12-in. steel channels weighing 25 
lb. /ft. are to be bolted to the sides of a wooden beam 4 
in. wide, 12 in. deep, and 20 ft. long. What uniform load 
including the weight of the beam may be carried ? Maxi- 
mum unit stress in the steel 12,000 Ib./sq. in., and 600 
Ib./sq. in. in the wood. 

In the following problems the maximum allowable unit 
stress in the concrete may be taken as 600 Ib./sq. in. and 
steel as 12,000 Ib./sq. in., and the value of y as 10. 

6. A reenforced concrete beam 6 in. deep, 4 ft. wide, 
and 6 ft. span has 2 sq. in. of steel reenforcement placed 
2 in. from the lower side. What uniform load including 



COMPOUND BARS AND BEAMS 175 

its own weight will it carry? What are the maximum 
unit stresses in the steel and the concrete ? 

7. Find the proper area for the steel in a beam similar 
to the one given in problem 6. 

8. A concrete beam 10 ft. long and 6 in. square is to be 
reenforced by steel rods placed 2 in. from the lower side. 
Find the proper area for the steel reenforcement. 

9. What uniform load will the beam in problem 8 carry 
with safety ? 

10. What load could be carried without any reenforce- 
ment, assuming the tensile stress in the concrete to be 150 
Ib./sq. in.? 



176 MECHANICS OF MATERIALS 

TABLES. — EXPLANATION OF 

Table I. Notation. 

The number following the description of each symbol refers to the 
article where the symbol was introduced. 

Table II. Fundamental Formulas. 

Table III. Derived Formulas. 

The numbers following each expression refer to the chapter and 
article in which the formula was derived. 

Table IV. Properties of Beams. 

The columns 1 and 2 give the relative strengths and stiffness of the 
various kinds of beams of the same length and shape. Columns 3 to 6 
are the expressions for Maximum Vertical Shear, Bending Moment, 
Unit Stress, and deflection 'of the various beams under uniform and 
single concentrated loads. Columns 7 and 8 give the values of a and /3 
for the various beams ; for a description of these symbols see : for a, 
Art. 48 ; ^, Art. 64. 

Table V. Constants of Materials. 

This table has been compiled solely for the use of the student in 
solving the problems in the text. As all the constants are liable to 
considerable variation, it should not be used in the design of a structure 
that is to be built. 

Table VL Properties of Sections. 

In the rectangular sections d is the dimension in the direction of 
bending. In the hollow sections d^ and h^ are the inside dimensions. 

Tables VII and VIII. Properties of I and Channel Beams. — 
Cambria Steel. 

These tables have been inserted for the convenience of the student. 
As every engineer should own some of the trade books giving the 
properties of the various steel shapes, he will prefer to get his data 
first hand. 

NOTATION. — TABLE I 

A area of cross section. 4. 

a a distance. 72. 

b breadth of a beam. 41. 

C, Cj, Cg, etc. constants of integration. 64. 



NOTATION. — TABLE I 177 

c distance from neutral axis to most distant fiber. 40. 

D internal diameter of a pipe. 24. 

D mean diameter of coils (helical springs only). 61. 

Z)j external diameter of thick pipe. 25. 

d diameter of rivets. 27. 

d diameter of bolts in shaft couplings. 59. 

d diameter of wire for helical springs. 61. 

d dej^th of a beam. 41. 

d^ a distance (reenf orced concrete only) . 88. 

E modulus of elasticity. 12. 

Et^ Ec modulus of elasticity for concrete and steel (reenf orced con- 
crete only). 88. 

e total deformation in length of bar. 11. 

e a distance (reenf orced concrete only). 88. 

F factor of safety. 21. 

F shearing modulus of elasticity. 56. 

f maximum deflection for a beam. 73. 

h distance bet\Yeen parallel axes. 74. 

/ rectangular moment of inertia. 40. 

J polar moment of inertia. 52. 

J' polar moment of inertia of bolts about shaft axis (shaft coup- 
lings only). 59. 

K total elastic resistance of a bar. 18. 

k elastic resistance of a material. 17. 

I a length. 11. 

M the bending moment. 40. 

il/j, il/g bending moments due to resultant couples. 64. 

AIj. moment of resistance (reeiiforced concrete only). 88. 

N number of revolutions per minute. 58. . 

iVp iVg, iVg bending moments at the supports in continuous beams. 
67. 

n number of concentrated loads on a beam. 64. 

n constant depending on kind of column. 73. 

n number of bolts in a shaft coupling. 59. 

P external force. 4. 

p pitch of rivets. 30. 

p a distance. 43. 

p arm of twisting moment. 52. 

R pressure per square unit. 24. 

r radius of curvature. 63. 

r radius of gyration. 73. 



178 MECHANICS OF MATERIALS 

Tij, Tig, T13 reactions at the supports of a beam. 36. 

5 unit stress, with subscripts t, c, and s for unit stress in tension, 

compression, and shear. 9. 
Sh horizontal unit shearing stress in beams. 81. 

Sjj, Sn maximum shearing and tensile unit stresses due to combined 

stresses. 80. 
t thickness. 24:. 

W weight of a box or beam. 23. 

W total uniform load on beam, may include weight of beam. 36. 
iv weight of a cubic unit of material. 23. 

w width of plate. 27. 

w uniform load on beam per linear unit. 36. 

X, y variable distances. 
X, y, z ratios (for concrete beams only). 88. 
a, (i material constants. 46, 64. 
<^ constant depending on material. 73. 

B angle of twist. 52. 

FUNDAMENTAL FORMULAS. — TABLE II 

Tension, Compression, and Shear 
(a) P = AS. Chap. I, Art. 4. 

Applies to all cases of uniformly distributed stress. 

Modulus of Elasticity for Tension and Compression 

(h) E = - = ^. Chap. I, Art. 11. 

e Ae 

Applies to all problems where the unit stress in tension or com- 
pression is within the elastic limit. 

Beams. — Vertical Shear 

(c) V=AS. Chap. Ill, Art. 39. 
True for all values of S. 

Bending Moment 

(d) M = — . Chap. Ill, Art. 40. 

c 

Applies to all problems where the value of S is within the elastic 
limit. 



DERIVED FORMULAS. — TABLE III 179 



Twisting Moment in Shafts 

<e) Pp = —. Chap. IV, Art. 53. 

c 

Applies to all problems where the value of S is within the elastic 
limit. 

Equation of Elastic Curve 

EI-^ = iM. Chap. V, Art. 6:]. 



DERIVED FORMULAS. — TABLE III 
Strength of Bars of Uniform Strength 

logio A = 0.434 - y + logio Aq. Chap. II, Art. 23. 
w 

Thickness of Steam and Water Pipes, Cylinders, etc. 

Thin pipes : 

Longitudinal ruptures. 

RD = 2 St. Chap. II, Art. 24. 

Circumferential ruptures. 

RD = 4: St. Chap. II, Art. 24. 

Thick pipes : 

Longitudinal ruptures. 

RD^ = 2 St. Chap. II, Art. 25. 

Strength of Riveted Joints 
Tension in plate. KP — ^O'^t = ^t- Chap. II, Art. 30. 

Shear on rivet. ^^ S^ = P,. Chap. II, Art. 30. 

Compression on rivet or plate. c-^tdSc = Pc Chap. II, Art. 30. 

Horse Power of Shafts 

H= ^^^^ (approx.) = _M^, Chap. IV, Art. 58. 
63,000^ ^^ ^ 63,000 c ^ 



180 MECHANICS OF MATERIALS 

Shaft Couplings 

Diameter of bolts. Pp = 7i^ S/'h (approx.). Chap. IV, Art. 59. 

•1 

Helical Springs 
Strength. p = l^Ss. 

Deflection. S = ^^ = %^. Chap. IV, Art. 61. 

-tcl^ Fa 

Continuous Beams. — Three-moment Equation 

^A + 2 iV^ (I, + h) + 7VV2 = - ^^Vi' + ^^2' . Chap. V, Art. 67. 



Long Columns 



Round ends. 



Euler. P = ^If^. Chap. VI, Art. 70. 

AS 
Rankine. P = — - . Chap. VI, Art. 73. 

l+<^lL 



Square ends. 

Euler. P = 4 ^^. Chap. VI, Art. 71. 

Rankine. P = ^^^ - Chap. VI, Art. 73. 

Round and square ends. 

Euler. P = ^ ^I^, Chap. VI, Art. 72, 

Rankine. P = ^"^'^^ . Chap. VI, Art. 73. 

I + ^IT^' 
9 r^ 

Combined Stresses 
Tension or compression with bending. 

S = ^+ — — — ., Chap. VII, Art. 76. 

)g E 

p ^^c 

or S = h - — (approx.). 



DERIVED FORMULAS. — TABLE III 



1«1 



Maxiiimm tension, compression, or shear. 
Tension or compression combined with shear. 



Max. shear. 
Max. tension or 
compression. 



Sp = ± VSs^ + i S-^. Chap. YII, Art. 80. 

Sn = lS± VS;^ + i S^. Chap. YII, Art. 80. 



Horizontal Shearing Stresses in Beams 

S^ = -^ CydA. Chap. VII, Art. 8L 

clxlb •Jy-i 



and 



Reenforced Concrete Beams 
M= StzbdMl--\ 

X 

3 



il/=,Se^7l-^']. 



> Chap. VII, Art. 88. 



18: 



MECHANICS OF MATERIALS 



> 






< 

O 

GO 

»— t 

H 
O 

Ph 



00 aj. 


CO 


CO 


00 


00 


iC 


CO 

r-l 


(M 

cs 

r— ( 


X 

CO 


t- s 


1-H 


CM 


Ti^ 


X 


X 


^ 


G 
Max. 

TION 




CO 


TO 


00 


CO 


30 


IS 

Id X) 

-TO 


TO 


:o 

X 

1— ( 


TO 


'M 

1— ( 


TO 


— r< 
X 

CO 


5 

Max. 

Tensile or 

('O.Ml'RKSSIVE 

Unit Stress 




^ 


^ 
S 


CM 


^ 

g 




O 

S 




o 

s 


X 


O 

5 


X 


o 

~-~i 

^ 


r— ( 


K S W 


""^ 

u 


g^l 




^ 00 


^ X 


^ X 


tS rH 


3 

Max. 

Vertical 

Shear 


^ 


^ 


^1^1 


klo. 




00 


^[CM 


^1^, 


2 

Relative 

Stiffness 


1— 1 


GOICO 


T— 1 


00 


CM 

CO 


"^i 

o 


X 

T— 1 


1 

Relative 
Strknutii 


tH 


C^ 


-* 


CX) 


X 


X 


<M 

1—1 


g 

o 

fi 
z 

s 

« 

o 

o 

z 




4-3 

o; 

ce 

o 

> 




la 

£ 
o 

*-( 

a; 

■C! 

o 

— ( 

^-( 

> 

ID 

5 
J 


1— ( 

s 

'Eh 

g 

cJ5 


o 

"S 

<D 

'Eh 

£ 


1e 

V 
•4-3 

o 

£h 

CO 

33 

<v 
c 
o 

-t-3 


a 
o 

o 


-4-3 

r3 

a 
o 
I— 1 

'^ 
<D 

-k3 

o 

■+3 

c3 
'IS 
X 


3 


1 

V 
O^ 
-^ 

-4J 

o 

-4-3 

>< 

5 

PQ 


5 

tH-l 



AVERAGE CONSTANTS OF MATERIALS. — TABLE V 183 



pq 



w 

P^ 
H 

o 

H 
H 

CO 

1^ 

O 
o 

o 

p^ 



ZD 
O 

a 

o 

b 


o 


T— 1 







J— 1 





10 
1—1 








c 




o 
1— ( 


I— 1 


CO 





OS 


10 

CI 


10 



CI 


11 


00 


00 


^ 


'^l 





1—1 




1— ( 


1— 1 


Shearing' 

Modulus of 

Elasticity 

Lbs./Sq. In. 


o 

cT 

T— 1 













10 







o~ 



0" 










oP 

T— ( 







o" 


0" 














Modulus of 
Elasticity 

Lbs./Sq. In. 


o 
o 
o 

o 

o 
o 
1—1 









or 

T— 1 






0" 



00" 

CI 









0" 

CO 






0" 





CO 
















o_ 

CO 


H 
■J 

H 

< 




■5 
■- 







o~ 




8 




0^ 


10 








05 











1— ( 












GO 









00 






Iff 
10 






0" 

CD 








T— * 







CO 






CD 






CO 


;i5 






00 







o~ 

OJ 











0" 










1— t 















Elastic 

Limit 

EL 

Lbs./Sq. In. 






CO 






CO 






10 

CM 






0" 

CO 







00 














IS 




u 

s 

■ iH 

H 




-(J 




p 

• 1-H 

bJO 

g 


u 


-1-3 

C/2 


-(-3 

00 

!-< 

hH 

HH 


c. 
PC 




a 
c 
c/: 


) 

i 

\ 


a 
a 

z 

c 





18i 



MECHANICS OF MATERIALS 



> 



O 

•—I 

H 
o 

::/2 



t— ( 

>-i 
O 
Ah 



o 

02 


^=1 O 


2 

+ 


+ 
> 








TO 1— 1 


1 

-r 




S| w 


"5 o 


2? I — ^ 






TO ^J 

1 1— ( 


1 




5 5 < 


<* 


^13^ 


« I'M 

^ 1 T— ( 


1 
eo 




£L 1 ^ 




1—1 


92 
X 

> 

H 

r- 

5 
t— 1 

"a 
O 

^: 


S 


+ 


I— 1 


^ 1 ^ 






^ d 
N ^ 


1 


(M 

CO 


<- 




^ Ir-I 


1 

TO 


T) 




1 




g 
-< 


"^ 


•^ 




1 


H 

"g-* 


1 

e-i 


TjH 


1 

o 

H 
OQ 

O 

a 

OQ 




If 

C/2 


r-cT 

3 


■02 


< 

1 

c 

-I- 

c 

c 

! 
\ 


5 

H 




( 
1 

( 

1 

1- 


3 

J 



INDEX 



Axial force, 3. 

Bar, A compound, 165, 

Definitiou of, 3. 

of uniform strength, 22, 179. 
Beams, Compound, 163. 

Continuous, 109, 180. 

Deflection of, 102. 

Kinds of, 46. 
_ Maximum stress in, 156. 

of uniform strength, 66. 

Overhanging, 65. 

Reactions at the supports of, 47. 

Reenforced concrete, 1(56, 181. 

Relative strengths of, 64. 

Restrained or fixed, 105, 106, 107. 
Bending moment, in beams, 50, 178. 

Relation between vertical shear 
and, 63. 

Columns, Long, 123. 

Euler's formula for, 125, 180. 

Rankine's formula for, 131, 180. 
Compound bars and beams, 163. 
Compression combined with shear, 149. 

Formula for, 3, 178. 
Concrete beams, Reenforced, 166, 181. 
Continuous beams, 109, 180. 

Dangerous section in beams, 63. 
Deformation, of elastic bodies, 7. 

Unit, 7. 
Ductility, 11. 

Elastic Limit, 8. 

Commercial, 9. 
Elastic curve, 48, 99. 

Equation of the, 99, 179. 
Elasticity, Modulus of, 8, 178. 
Euler's formula for long columns, 125. 

Factors of safety, 14. 

Force, Internal and external, 6. 



Formula, Table of fundamental, 178. 
Table of derived, 179. 

Helical springs, 93. 
Horizontal shear in beams, 152. 

Load, Concentrated or uniform — on 
beams, 47. 
Eccentric axial — on beams, 148. 
Moving — on beams, 67. 

Materials, Constants of, 13, 183. 
Modulus of. 

Elasticity, Tension or compression, 
8. 

Elasticity, Flexure, 102. 

Elasticity, Torsion, 88, 183. 

Rupture, Tension or compression, 3. 

Rupture, for beams, 55. 

Rupture, for torsion, 92. 

Section, Beams, 56. 

Section, Torsion, 86. 

for beams, 56. 
Moment, Bending, 50, 178. 

Diagram, 58. 

Resisting, 52. 

Neutral axis or plane, 53. 
Notation, Table of, 176. 

Pipes, Thin, strength of, 25, 179. 
Thick, strength of, 179, 27. 

Rankine's formula for long columns, 

131. 
Reactions for beams, 47. 
Reenforced concrete beams, 166. 
Resilience, 10. 

Ultimate, 11. 

Elastic, 12. 
Riveted joints, .30. 

Compression in, 33, 179. 

Etiiciency of, 37. 



185 



186 



INDEX 



Riveted joints {continued) 
General case of, 06. 
Kinds of, 35. 
Slieai- in, 32, 179. 
Tension in, 30, 179. 

Sections, Pi-operties of, 18i. 

Square — in Torsion, 8G. 
Shafts, Couplings for, 90, 180. 

Horse power of, 89, 179. 

Strength and stiffness of, 89. 

Twist of, 88. 

Twisting moment in, 179. 
Shear, and axial stress, 149. 

Diagrams, 58, 59. 

Horizontal shear inheams, 152, 181. 

Resisting, 51. 

Vertical — in beams, 49, 179. 
Shearing stress, 5. 
Springs, Helical, 93, 180. 
Strength, Bars of uniform, 22, 179. 

of cylinders, pipes, and spheres, 25. 

of thick pipes, 27. 

Ultimate — of materials, 9. 
Stress, 3. 
Stress, Combined, 143, 180, 181. 

in roof rafters, 147. 



in long columns, 123. 
due to change of temperature, 39. 
Maximum — in beams, G3, 150. 
Maximum — tensile or compressive, 

5, 178. 
Shearing, 5, 178. 
Tensile or compressive, 4. 
Unit, 4. 
Working, 14. 

Tables : Constants of materials, 183. 

Derived formulas, 179. 

Fundamental formulas, 178. 

Notation, 176. 

Properties of beams, 182. 

Properties of sections, 184. 
Tension, combined with bending, 144. 

combined with shear, 149. 

Formula for, 3, 178. 
Torsion, Derivation of formula for, 83. 

Modulus of elasticity for, 183. 

Vertical shear, 49. 
Relation between the bending mo- 
ment and, 63. 

Yield point, 9. 



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